Explanation:
Okay, so I need to find the minimum value of the expression $x^4 + 4y^2 + 4z^4$ given that $xyz = 2$ where $x$, $y$, and $z$ are positive real numbers. Hmm, let me think about how to approach this.

First, since there's a constraint $xyz = 2$, maybe I can use the method of Lagrange multipliers? Or perhaps there's a way to apply the AM-GM inequality here? I remember that AM-GM is useful for optimization problems with products and sums. Let me recall the AM-GM inequality: for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. So maybe I can split the terms in the expression into parts that can be related to the constraint.

Looking at the expression $x^4 + 4y^2 + 4z^4$, I notice that there are different exponents: 4, 2, and 4. The constraint is a product of variables to the first power. Maybe I need to adjust the terms so that their exponents can be matched with the constraint somehow.

Alternatively, maybe I can use substitution. Since $xyz = 2$, perhaps I can express one variable in terms of the other two. Let me try that. For example, solve for $y$: $y = \frac{2}{xz}$. Then substitute this into the original expression:

$x^4 + 4\left(\frac{2}{xz}\right)^2 + 4z^4 = x^4 + 4\left(\frac{4}{x^2 z^2}\right) + 4z^4 = x^4 + \frac{16}{x^2 z^2} + 4z^4$.

Hmm, now I have an expression in terms of $x$ and $z$. But this seems more complicated. Maybe this isn't the right path. Let me think again.

Alternatively, maybe I can apply AM-GM directly to the terms $x^4$, $4y^2$, and $4z^4$. Wait, but AM-GM requires the terms to be positive, which they are. However, the problem is that the number of terms isn't clear. If I consider $x^4$ as one term, $4y^2$ as another, and $4z^4$ as the third, but that might not be helpful. Let me check.

Alternatively, maybe I can split $x^4$ into multiple terms so that I can apply AM-GM. For example, split $x^4$ into four terms each of $x^4/4$, and similarly split $4y^2$ into two terms of $2y^2$, and $4z^4$ into four terms of $z^4$. Wait, let me think. If I split the expression into more terms, I can apply AM-GM over those terms. Let me try:

Let me write the expression as:

$x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4$.

Wait, that would be splitting $4y^2$ into two terms of $2y^2$ and $4z^4$ into four terms of $z^4$. Wait, but then the total number of terms would be 1 (from x^4) + 2 (from 4y^2) + 4 (from 4z^4) = 7 terms. Then, applying AM-GM:

$$
\frac{x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4}{7} \geq \sqrt[7]{x^4 \cdot (2y^2)^2 \cdot (z^4)^4}
$$

Wait, let me check that. If I have seven terms: one term is $x^4$, two terms are $2y^2$, and four terms are $z^4$. Wait, but that would be:

$ x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4 $.

But the original expression is $x^4 + 4y^2 + 4z^4$, which is equal to the above. So, if I take the arithmetic mean of these seven terms:

$$
\frac{x^4 + 4y^2 + 4z^4}{7} \geq \sqrt[7]{x^4 \cdot (2y^2)^2 \cdot (z^4)^4}
$$

Wait, but hold on, the product inside the root would be the product of all seven terms. However, if I split the original expression into seven terms, then the product of those terms is $x^4 \times (2y^2)^2 \times (z^4)^4$? Wait, but the original expression isn't split like that. Wait, actually, the original expression is $x^4 + 4y^2 + 4z^4$, which is equal to $x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4$. So, if I take the product of those seven terms, it's $x^4 \times (2y^2)^2 \times (z^4)^4$? Wait, no, because the first term is x^4, then two terms of 2y^2, and four terms of z^4. Therefore, the product is:

$x^4 \times (2y^2)^2 \times (z^4)^4$. Let me compute that:

$x^4 \times 4y^4 \times z^{16}$. So:

$4x^4 y^4 z^{16}$.

But our constraint is $xyz = 2$, so maybe we can express $x^4 y^4 z^{16}$ in terms of that?

Let me compute $x^4 y^4 z^{16} = (xyz)^4 \times z^{8}$. Wait, since $xyz = 2$, then $(xyz)^4 = 16$, so $x^4 y^4 z^{16} = 16 z^{8}$. Therefore, the product inside the root is $4 \times 16 z^8 = 64 z^8$. Wait, but this seems messy, since z is still in there. So the geometric mean would be $\sqrt[7]{64 z^8}$, which depends on z. But since z is a variable, this approach might not give me a constant lower bound. Therefore, maybe my splitting isn't correct?

Alternatively, perhaps I need to adjust how I split the terms so that the product relates to the constraint $xyz = 2$.

Let me think again. Maybe I should split the terms in such a way that when I take the product, it's a constant. For example, if I can write the product of the terms in the AM-GM as something involving $xyz$. Let me try again.

Suppose I want to split the expression $x^4 + 4y^2 + 4z^4$ into terms where the product is a constant. Let me think about exponents. Let me try to use weighted AM-GM. The weighted AM-GM inequality states that for positive real numbers and positive weights that sum to 1, the weighted arithmetic mean is greater than or equal to the weighted geometric mean.

Alternatively, maybe I can use the standard AM-GM with multiple variables. Let me consider splitting the terms so that each term in the AM-GM corresponds to different variables. Let me see.

Suppose I split the expression into parts that can be related to the constraint. Let me think of the exponents.

The constraint is $xyz = 2$, which is a product of variables to the first power. The given expression is $x^4 + 4y^2 + 4z^4$. Maybe I need to balance the exponents.

Alternatively, maybe I can use substitution variables. For example, set $a = x^2$, $b = y$, and $c = z^2$. Then, the constraint becomes $x y z = 2$, which would be $\sqrt{a} \cdot b \cdot \sqrt{c} = 2$, so $\sqrt{a c} \cdot b = 2$. The expression becomes $a^2 + 4b^2 + 4c^2$. Hmm, not sure if that helps. Let me check:

Original expression: $x^4 + 4y^2 + 4z^4 = a^2 + 4b^2 + 4c^2$. The constraint is $\sqrt{a} \cdot b \cdot \sqrt{c} = 2$, which is equivalent to $\sqrt{a c} \cdot b = 2$. Maybe that's still complicated. Perhaps not helpful. Let me think of another way.

Alternatively, use Lagrange multipliers. Let me try that method.

Define the function to minimize as $f(x, y, z) = x^4 + 4y^2 + 4z^4$, subject to the constraint $g(x, y, z) = xyz - 2 = 0$.

The method of Lagrange multipliers says that at the minimum, the gradient of f is proportional to the gradient of g. Therefore:

$$
\nabla f = \lambda \nabla g
$$

Compute the gradients:

Gradient of f:

$$
\frac{\partial f}{\partial x} = 4x^3, \quad \frac{\partial f}{\partial y} = 8y, \quad \frac{\partial f}{\partial z} = 16z^3
$$

Gradient of g:

$$
\frac{\partial g}{\partial x} = yz, \quad \frac{\partial g}{\partial y} = xz, \quad \frac{\partial g}{\partial z} = xy
$$

Therefore, setting up the equations:

1. $4x^3 = \lambda yz$
2. $8y = \lambda xz$
3. $16z^3 = \lambda xy$
4. $xyz = 2$

So, we have four equations. Let me try to solve them.

From equations 1, 2, 3, we can express $\lambda$ in terms of variables and set them equal.

From equation 1: $\lambda = \frac{4x^3}{yz}$

From equation 2: $\lambda = \frac{8y}{xz}$

Set them equal:

$$
\frac{4x^3}{yz} = \frac{8y}{xz}
$$

Multiply both sides by $yz \cdot xz$ to eliminate denominators:

$4x^3 \cdot xz = 8y \cdot yz$

Simplify:

$4x^4 z = 8y^2 z$

Divide both sides by 4z (since z is positive, so z ≠ 0):

$x^4 = 2y^2$

So, $x^4 = 2y^2$, which implies $y^2 = \frac{x^4}{2}$, so $y = \frac{x^2}{\sqrt{2}}$. Since y is positive.

Similarly, let's take equations 2 and 3:

From equation 2: $\lambda = \frac{8y}{xz}$

From equation 3: $\lambda = \frac{16z^3}{xy}$

Set them equal:

$$
\frac{8y}{xz} = \frac{16z^3}{xy}
$$

Multiply both sides by $xz \cdot xy$:

$8y \cdot xy = 16z^3 \cdot xz$

Wait, let me check:

Left side: $\frac{8y}{xz} \cdot xz \cdot xy = 8y \cdot xy = 8x y^2$

Right side: $\frac{16z^3}{xy} \cdot xz \cdot xy = 16z^3 \cdot xz = 16x z^4$

Therefore:

$8x y^2 = 16x z^4$

Divide both sides by 8x (since x is positive):

$y^2 = 2 z^4$

Therefore, $y^2 = 2 z^4$, so $y = \sqrt{2} z^2$

Earlier, we had $y = \frac{x^2}{\sqrt{2}}$. Therefore, equate the two expressions for y:

$$
\frac{x^2}{\sqrt{2}} = \sqrt{2} z^2
$$

Multiply both sides by $\sqrt{2}$:

$$
x^2 = 2 z^2
$$

Therefore, $x^2 = 2 z^2$, so $x = \sqrt{2} z$ (since x, z are positive)

So now, we have relations between x, y, z:

From above:

1. $x = \sqrt{2} z$
2. $y = \sqrt{2} z^2$
3. Also, from $x^4 = 2 y^2$, let me check if they are consistent.

Let me check with equation 1: $x = \sqrt{2} z$, so $x^4 = (\sqrt{2} z)^4 = (2^{1/2})^4 z^4 = 2^{2} z^4 = 4 z^4$

And $2 y^2 = 2 (\sqrt{2} z^2)^2 = 2 (2 z^4) = 4 z^4$. So that's consistent.

Similarly, from equation $y^2 = 2 z^4$, so $y = \sqrt{2} z^2$, which is consistent.

So now, we can express all variables in terms of z. Let me do that.

Let me write:

x = √2 z

y = √2 z²

We also have the constraint xyz = 2. Let me substitute these into the constraint.

xyz = (√2 z) * (√2 z²) * z = Let me compute step by step.

First, √2 * √2 = 2

Then, z * z² * z = z^{1 + 2 + 1} = z^4

Therefore, xyz = 2 * z^4 = 2

Therefore:

2 z^4 = 2 => z^4 = 1 => z = 1 (since z is positive)

Therefore, z = 1.

Thus, x = √2 * 1 = √2

y = √2 * 1² = √2

So, x = √2, y = √2, z = 1

Let me check the constraint: xyz = √2 * √2 * 1 = (√2)^2 * 1 = 2 * 1 = 2. Correct.

Now, let's compute the value of the expression $x^4 + 4y^2 + 4z^4$.

Compute each term:

x^4 = (√2)^4 = (2^{1/2})^4 = 2^{2} = 4

4y^2 = 4*(√2)^2 = 4*2 = 8

4z^4 = 4*(1)^4 = 4*1 = 4

So total expression: 4 + 8 + 4 = 16.

Therefore, according to this, the minimum value is 16?

Wait, but let me check if this is indeed the minimum. Maybe there's a mistake here? Let me verify.

Alternatively, maybe I can test another set of values. Suppose I take x, y, z such that xyz = 2. For example, take x = 2, y = 1, z = 1. Then xyz = 2*1*1 = 2. Compute the expression: x^4 + 4y^2 + 4z^4 = 16 + 4 + 4 = 24, which is higher than 16. So 16 seems better.

Another test case: x = 1, y = 2, z = 1. Then xyz = 1*2*1 = 2. The expression is 1 + 4*(4) + 4 = 1 + 16 + 4 = 21, still higher.

Another test case: Let me try x = √2, y = √2, z = 1. Which is the critical point we found. So 4 + 8 + 4 = 16.

What if I take another critical point? Wait, how do I know that this is the only critical point? Well, since we used Lagrange multipliers and found this solution, and the function is convex? Not sure. But given that the test cases give higher values, maybe 16 is indeed the minimum. Let me check if there's a way to confirm using AM-GM.

Earlier, when I tried splitting the terms, perhaps I messed up the splitting. Let me try again. Let me think of splitting the expression into terms such that when applying AM-GM, the product relates to the constraint.

We have the expression x^4 + 4y^2 + 4z^4. Let me consider the terms as follows:

Let me think of x^4 as one term, 4y^2 as two terms of 2y^2 each, and 4z^4 as four terms of z^4 each. Wait, this is similar to before, but maybe I can use weighted AM-GM. Alternatively, think of it as:

Let me write the expression as x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4. That is, 1 + 2 + 4 terms. Wait, total of 1 + 2 + 4 = 7 terms. Then, applying AM-GM:

( x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4 ) / 7 ≥ ( x^4 * (2y^2)^2 * (z^4)^4 )^{1/7}

Wait, compute the product inside:

x^4 * (2y^2)^2 * (z^4)^4 = x^4 * 4y^4 * z^{16} = 4 x^4 y^4 z^{16}

But from the constraint, xyz = 2, so xyz = 2. Let me express x^4 y^4 z^{16} in terms of (xyz)^something.

Note that x^4 y^4 z^{16} = (xyz)^4 * z^{8} = (2)^4 * z^8 = 16 z^8

So the product is 4*16 z^8 = 64 z^8. Therefore:

AM-GM gives:

[Expression]/7 ≥ (64 z^8)^{1/7}

But this is still dependent on z, so not helpful. Therefore, this approach might not give a constant lower bound.

Alternatively, perhaps I need to split into different numbers of terms. Maybe using different coefficients.

Alternatively, think about the expression as x^4 + 4y^2 + 4z^4 and consider variables substitution. Let me set variables to make exponents match.

Let me denote a = x^2, b = y, c = z^2. Then, the original expression becomes:

a^2 + 4b^2 + 4c^2.

The constraint is xyz = 2. Since x = sqrt(a), y = b, z = sqrt(c), then the constraint is sqrt(a) * b * sqrt(c) = 2, which is sqrt(a c) * b = 2. So, b sqrt(a c) = 2.

So we need to minimize a^2 + 4b^2 + 4c^2 with b sqrt(a c) = 2.

This seems a bit complex, but maybe we can apply Cauchy-Schwarz or AM-GM here.

Alternatively, let me try substitution. From the constraint: b = 2 / sqrt(a c). Substitute into the expression:

Expression becomes:

a^2 + 4*(2 / sqrt(a c))^2 + 4c^2 = a^2 + 4*(4 / (a c)) + 4c^2 = a^2 + 16/(a c) + 4c^2.

Now, we have variables a and c. Maybe we can set t = a c, but not sure. Alternatively, set variables such that we can apply AM-GM.

Alternatively, think of a^2 + 4c^2 as two terms and 16/(a c) as another term. Maybe apply AM-GM on three terms? Let me try:

Let me consider the three terms: a^2, 4c^2, and 16/(a c). Wait, but they are not symmetric. However, maybe I can adjust them.

Alternatively, think of splitting 16/(a c) into multiple terms. Let me think of splitting the expression into more terms.

Let me try:

Let me think of the expression as a^2 + 4c^2 + 16/(a c). Maybe apply AM-GM on three terms. Let me check:

By AM-GM:

(a^2 + 4c^2 + 16/(a c))/3 ≥ cube root( a^2 * 4c^2 * 16/(a c) )

Compute the product inside:

a^2 * 4c^2 * 16/(a c) = 4 * 16 * a^2 c^2 / (a c) ) = 64 a c.

Therefore, the cube root of 64 a c = cube root(64) * cube root(a c) = 4 * cube root(a c).

So, the inequality becomes:

(a^2 + 4c^2 + 16/(a c))/3 ≥ 4 * cube root(a c)

But this seems messy, since we still have cube root(a c). However, if we can relate a c to something. From the constraint, earlier, we had that b = 2 / sqrt(a c). But in this substitution, I don't think that connects directly. Alternatively, perhaps there's another way.

Alternatively, set variables such that a c is fixed? Not sure. Maybe this approach isn't the best.

Alternatively, go back to the previous result from Lagrange multipliers. Since we found that the critical point gives the value 16, and testing other points gives higher values, maybe 16 is indeed the minimum.

But let me check if there is another way to apply AM-GM. Let me think again.

Original expression: x^4 + 4y^2 + 4z^4.

Let me think of the terms as follows: x^4, 2y^2, 2y^2, z^4, z^4, z^4, z^4. As before, 7 terms. Then, applying AM-GM:

( x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4 ) / 7 ≥ (x^4 * (2y^2)^2 * (z^4)^4 )^{1/7}

Which simplifies to:

[Expression]/7 ≥ [x^4 * 4y^4 * z^{16}]^{1/7} = [4x^4 y^4 z^{16}]^{1/7}

But since xyz = 2, then x^4 y^4 z^{16} = (xyz)^4 * z^8 = 2^4 * z^8 = 16 z^8.

Thus, the right side is [4 * 16 z^8]^{1/7} = [64 z^8]^{1/7} = 64^{1/7} * z^{8/7}

But since z is a variable, unless z is fixed, this won't help. However, if we can relate z to something else. From the critical point found earlier, z = 1. If we assume that the minimum occurs when z = 1, then maybe?

Alternatively, perhaps there's an error in splitting the terms. Maybe I need to split the expression into different numbers of terms. Let me try another approach.

Suppose I use AM-GM on x^4 and 4z^4. Let me consider x^4 and 4z^4 as two terms. Then, by AM-GM:

(x^4 + 4z^4)/2 ≥ sqrt(x^4 * 4z^4) = sqrt(4x^4 z^4) = 2x^2 z^2.

Therefore, x^4 + 4z^4 ≥ 4x^2 z^2.

Similarly, for the 4y^2 term, maybe pair with something. But we have 4y^2. If we consider 4y^2, maybe combine with other terms?

Alternatively, if I consider the entire expression:

x^4 + 4y^2 + 4z^4 ≥ 4x^2 z^2 + 4y^2.

But then, we can apply AM-GM on 4x^2 z^2 and 4y^2. Let me consider them as two terms:

(4x^2 z^2 + 4y^2)/2 ≥ sqrt(4x^2 z^2 * 4y^2) = sqrt(16x^2 y^2 z^2) = 4xyz.

Therefore, 4x^2 z^2 + 4y^2 ≥ 8xyz.

Therefore, the original expression:

x^4 + 4y^2 + 4z^4 ≥ 4x^2 z^2 + 4y^2 ≥ 8xyz.

Given that xyz = 2, so this would give:

Original expression ≥ 8 * 2 = 16.

Therefore, the minimum value is at least 16. And since we found a case where the expression equals 16, which satisfies the equality condition, that must be the minimum.

Wait, but how does the equality condition hold?

For AM-GM equality, all terms must be equal. So in the first step, when we applied AM-GM to x^4 and 4z^4, we have equality when x^4 = 4z^4 → x^2 = 2 z^2 → x = √2 z (since positive). Then, for the second AM-GM step, we applied to 4x^2 z^2 and 4y^2, so equality when 4x^2 z^2 = 4y^2 → x^2 z^2 = y^2 → y = x z. But from earlier, we had x = √2 z, so substituting:

y = x z = √2 z * z = √2 z^2. Which is exactly the relations found via Lagrange multipliers. Then, with xyz = 2, substituting x = √2 z and y = √2 z^2, we get:

xyz = √2 z * √2 z^2 * z = 2 z^4 = 2 → z^4 = 1 → z = 1. Therefore, all equality conditions are satisfied. Therefore, the minimum value is indeed 16. So this confirms the result from Lagrange multipliers.

Therefore, the minimum value of the expression is 16.

I think this is solid. The AM-GM approach gives the lower bound of 16, and we have a point where equality is achieved, so that must be the minimum. So the answer is 16.

**Final Answer**
\boxed{16}

To find the minimum value of the expression
$$
x^4 + 4y^2 + 4z^4
$$
under the constraint
$$
xyz = 2, \quad \text{where } x, y, z > 0,
$$
we consider the application of the **AM-GM Inequality** and the **Method of Lagrange Multipliers**.

---

### Step 1: Apply the AM-GM Inequality

We observe the expression and consider a modified form of the AM-GM inequality. Let us consider the following breakdown of the expression:

$$
x^4 + 4y^2 + 4z^4 = x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4
$$

This partitions the expression into **7 terms**:
- One term of $ x^4 $,
- Two terms of $ 2y^2 $, and
- Four terms of $ z^4 $.

Now, applying the AM-GM inequality to these 7 terms:

$$
\frac{x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4}{7} \geq \sqrt[7]{x^4 \cdot (2y^2)^2 \cdot (z^4)^4}
$$

Simplifying the right-hand side:

$$
\sqrt[7]{x^4 \cdot 4y^4 \cdot z^{16}} = \sqrt[7]{4x^4 y^4 z^{16}}
$$

But $ xyz = 2 $, so
$$
x^4 y^4 z^{16} = (xyz)^4 \cdot z^8 = 2^4 \cdot z^8 = 16z^8
$$

So the inequality becomes:
$$
\frac{x^4 + 4y^2 + 4z^4}{7} \geq \sqrt[7]{4 \cdot 16z^8} = \sqrt[7]{64z^8}
$$

Thus,
$$
x^4 + 4y^2 + 4z^4 \geq 7 \cdot \sqrt[7]{64z^8}
$$

But this result still depends on $ z $. To find a **constant** lower bound, we must ensure that the product inside the root is a constant. This suggests more symmetric reasoning is necessary.

---

### Step 2: Apply AM-GM on Two Pairs

Let us now try splitting the expression into two parts. Consider the inequality:
$$
x^4 + 4z^4 \geq 2\sqrt{x^4 \cdot 4z^4} = 2\sqrt{4x^4 z^4} = 4x^2 z^2
$$

So we have:
$$
x^4 + 4y^2 + 4z^4 \geq 4x^2 z^2 + 4y^2
$$

Now, apply AM-GM to the two terms $ 4x^2 z^2 $ and $ 4y^2 $:
$$
\frac{4x^2 z^2 + 4y^2}{2} \geq \sqrt{4x^2 z^2 \cdot 4y^2} = \sqrt{16x^2 y^2 z^2} = 4xyz
$$

Thus:
$$
4x^2 z^2 + 4y^2 \geq 8xyz
$$

Therefore:
$$
x^4 + 4y^2 + 4z^4 \geq 8xyz = 8 \cdot 2 = 16
$$

This gives the **lower bound** of **16**.

---

### Step 3: Check for Equality

To confirm that the **minimum value of 16** is achievable, we find values of $ x, y, z $ such that all AM-GM equalities are satisfied. That happens when:

- $ x^4 = 4z^4 $ → $ x^2 = 2z^2 $ → $ x = \sqrt{2}z $
- $ y^2 = x^2 z^2 = 2z^4 $ → $ y = \sqrt{2}z^2 $

So, substituting back into the constraint $ xyz = 2 $:
$$
x = \sqrt{2}z, \quad y = \sqrt{2}z^2 \Rightarrow xyz = \sqrt{2}z \cdot \sqrt{2}z^2 \cdot z = 2z^4 = 2 \Rightarrow z^4 = 1 \Rightarrow z = 1
$$

Hence:
- $ x = \sqrt{2} $
- $ y = \sqrt{2} $
- $ z = 1 $

Now, substitute these into the original expression:

$$
x^4 + 4y^2 + 4z^4 = (\sqrt{2})^4 + 4(\sqrt{2})^2 + 4(1)^4 = 4 + 8 + 4 = 16
$$

So, the **minimum value is indeed achieved** at $ x = \sqrt{2}, y = \sqrt{2}, z = 1 $, and the minimum value of the expression is:

$$
\boxed{16}
$$

[Assistant]
Explanation:
Okay, so I need to find the minimum value of the expression $x^4 + 4y^2 + 4z^4$ given that $xyz = 2$ where $x$, $y$, and $z$ are positive real numbers. Hmm, let me think about how to approach this.

First, since there's a constraint $xyz = 2$, maybe I can use the method of Lagrange multipliers? Or perhaps there's a way to apply the AM-GM inequality here? I remember that AM-GM is useful for optimization problems with products and sums. Let me recall the AM-GM inequality: for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. So maybe I can split the terms in the expression into parts that can be related to the constraint.

Looking at the expression $x^4 + 4y^2 + 4z^4$, I notice that there are different exponents: 4, 2, and 4. The constraint is a product of variables to the first power. Maybe I need to adjust the terms so that their exponents can be matched with the constraint somehow.

Alternatively, maybe I can use substitution. Since $xyz = 2$, perhaps I can express one variable in terms of the other two. Let me try that. For example, solve for $y$: $y = \frac{2}{xz}$. Then substitute this into the original expression:

$x^4 + 4\left(\frac{2}{xz}\right)^2 + 4z^4 = x^4 + 4\left(\frac{4}{x^2 z^2}\right) + 4z^4 = x^4 + \frac{16}{x^2 z^2} + 4z^4$.

Hmm, now I have an expression in terms of $x$ and $z$. But this seems more complicated. Maybe this isn't the right path. Let me think again.

Alternatively, maybe I can apply AM-GM directly to the terms $x^4$, $4y^2$, and $4z^4$. Wait, but AM-GM requires the terms to be positive, which they are. However, the problem is that the number of terms isn't clear. If I consider $x^4$ as one term, $4y^2$ as another, and $4z^4$ as the third, but that might not be helpful. Let me check.

Alternatively, maybe I can split $x^4$ into multiple terms so that I can apply AM-GM. For example, split $x^4$ into four terms each of $x^4/4$, and similarly split $4y^2$ into two terms of $2y^2$, and $4z^4$ into four terms of $z^4$. Wait, let me think. If I split the expression into more terms, I can apply AM-GM over those terms. Let me try:

Let me write the expression as:

$x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4$.

Wait, that would be splitting $4y^2$ into two terms of $2y^2$ and $4z^4$ into four terms of $z^4$. Wait, but then the total number of terms would be 1 (from x^4) + 2 (from 4y^2) + 4 (from 4z^4) = 7 terms. Then, applying AM-GM:

$$
\frac{x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4}{7} \geq \sqrt[7]{x^4 \cdot (2y^2)^2 \cdot (z^4)^4}
$$

Wait, let me check that. If I have seven terms: one term is $x^4$, two terms are $2y^2$, and four terms are $z^4$. Wait, but that would be:

$ x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4 $.

But the original expression is $x^4 + 4y^2 + 4z^4$, which is equal to the above. So, if I take the arithmetic mean of these seven terms:

$$
\frac{x^4 + 4y^2 + 4z^4}{7} \geq \sqrt[7]{x^4 \cdot (2y^2)^2 \cdot (z^4)^4}
$$

Wait, but hold on, the product inside the root would be the product of all seven terms. However, if I split the original expression into seven terms, then the product of those terms is $x^4 \times (2y^2)^2 \times (z^4)^4$? Wait, but the original expression isn't split like that. Wait, actually, the original expression is $x^4 + 4y^2 + 4z^4$, which is equal to $x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4$. So, if I take the product of those seven terms, it's $x^4 \times (2y^2)^2 \times (z^4)^4$? Wait, no, because the first term is x^4, then two terms of 2y^2, and four terms of z^4. Therefore, the product is:

$x^4 \times (2y^2)^2 \times (z^4)^4$. Let me compute that:

$x^4 \times 4y^4 \times z^{16}$. So:

$4x^4 y^4 z^{16}$.

But our constraint is $xyz = 2$, so maybe we can express $x^4 y^4 z^{16}$ in terms of that?

Let me compute $x^4 y^4 z^{16} = (xyz)^4 \times z^{8}$. Wait, since $xyz = 2$, then $(xyz)^4 = 16$, so $x^4 y^4 z^{16} = 16 z^{8}$. Therefore, the product inside the root is $4 \times 16 z^8 = 64 z^8$. Wait, but this seems messy, since z is still in there. So the geometric mean would be $\sqrt[7]{64 z^8}$, which depends on z. But since z is a variable, this approach might not give me a constant lower bound. Therefore, maybe my splitting isn't correct?

Alternatively, perhaps I need to adjust how I split the terms so that the product relates to the constraint $xyz = 2$.

Let me think again. Maybe I should split the terms in such a way that when I take the product, it's a constant. For example, if I can write the product of the terms in the AM-GM as something involving $xyz$. Let me try again.

Suppose I want to split the expression $x^4 + 4y^2 + 4z^4$ into terms where the product is a constant. Let me think about exponents. Let me try to use weighted AM-GM. The weighted AM-GM inequality states that for positive real numbers and positive weights that sum to 1, the weighted arithmetic mean is greater than or equal to the weighted geometric mean.

Alternatively, maybe I can use the standard AM-GM with multiple variables. Let me consider splitting the terms so that each term in the AM-GM corresponds to different variables. Let me see.

Suppose I split the expression into parts that can be related to the constraint. Let me think of the exponents.

The constraint is $xyz = 2$, which is a product of variables to the first power. The given expression is $x^4 + 4y^2 + 4z^4$. Maybe I need to balance the exponents.

Alternatively, maybe I can use substitution variables. For example, set $a = x^2$, $b = y$, and $c = z^2$. Then, the constraint becomes $x y z = 2$, which would be $\sqrt{a} \cdot b \cdot \sqrt{c} = 2$, so $\sqrt{a c} \cdot b = 2$. The expression becomes $a^2 + 4b^2 + 4c^2$. Hmm, not sure if that helps. Let me check:

Original expression: $x^4 + 4y^2 + 4z^4 = a^2 + 4b^2 + 4c^2$. The constraint is $\sqrt{a} \cdot b \cdot \sqrt{c} = 2$, which is equivalent to $\sqrt{a c} \cdot b = 2$. Maybe that's still complicated. Perhaps not helpful. Let me think of another way.

Alternatively, use Lagrange multipliers. Let me try that method.

Define the function to minimize as $f(x, y, z) = x^4 + 4y^2 + 4z^4$, subject to the constraint $g(x, y, z) = xyz - 2 = 0$.

The method of Lagrange multipliers says that at the minimum, the gradient of f is proportional to the gradient of g. Therefore:

$$
\nabla f = \lambda \nabla g
$$

Compute the gradients:

Gradient of f:

$$
\frac{\partial f}{\partial x} = 4x^3, \quad \frac{\partial f}{\partial y} = 8y, \quad \frac{\partial f}{\partial z} = 16z^3
$$

Gradient of g:

$$
\frac{\partial g}{\partial x} = yz, \quad \frac{\partial g}{\partial y} = xz, \quad \frac{\partial g}{\partial z} = xy
$$

Therefore, setting up the equations:

1. $4x^3 = \lambda yz$
2. $8y = \lambda xz$
3. $16z^3 = \lambda xy$
4. $xyz = 2$

So, we have four equations. Let me try to solve them.

From equations 1, 2, 3, we can express $\lambda$ in terms of variables and set them equal.

From equation 1: $\lambda = \frac{4x^3}{yz}$

From equation 2: $\lambda = \frac{8y}{xz}$

Set them equal:

$$
\frac{4x^3}{yz} = \frac{8y}{xz}
$$

Multiply both sides by $yz \cdot xz$ to eliminate denominators:

$4x^3 \cdot xz = 8y \cdot yz$

Simplify:

$4x^4 z = 8y^2 z$

Divide both sides by 4z (since z is positive, so z ≠ 0):

$x^4 = 2y^2$

So, $x^4 = 2y^2$, which implies $y^2 = \frac{x^4}{2}$, so $y = \frac{x^2}{\sqrt{2}}$. Since y is positive.

Similarly, let's take equations 2 and 3:

From equation 2: $\lambda = \frac{8y}{xz}$

From equation 3: $\lambda = \frac{16z^3}{xy}$

Set them equal:

$$
\frac{8y}{xz} = \frac{16z^3}{xy}
$$

Multiply both sides by $xz \cdot xy$:

$8y \cdot xy = 16z^3 \cdot xz$

Wait, let me check:

Left side: $\frac{8y}{xz} \cdot xz \cdot xy = 8y \cdot xy = 8x y^2$

Right side: $\frac{16z^3}{xy} \cdot xz \cdot xy = 16z^3 \cdot xz = 16x z^4$

Therefore:

$8x y^2 = 16x z^4$

Divide both sides by 8x (since x is positive):

$y^2 = 2 z^4$

Therefore, $y^2 = 2 z^4$, so $y = \sqrt{2} z^2$

Earlier, we had $y = \frac{x^2}{\sqrt{2}}$. Therefore, equate the two expressions for y:

$$
\frac{x^2}{\sqrt{2}} = \sqrt{2} z^2
$$

Multiply both sides by $\sqrt{2}$:

$$
x^2 = 2 z^2
$$

Therefore, $x^2 = 2 z^2$, so $x = \sqrt{2} z$ (since x, z are positive)

So now, we have relations between x, y, z:

From above:

1. $x = \sqrt{2} z$
2. $y = \sqrt{2} z^2$
3. Also, from $x^4 = 2 y^2$, let me check if they are consistent.

Let me check with equation 1: $x = \sqrt{2} z$, so $x^4 = (\sqrt{2} z)^4 = (2^{1/2})^4 z^4 = 2^{2} z^4 = 4 z^4$

And $2 y^2 = 2 (\sqrt{2} z^2)^2 = 2 (2 z^4) = 4 z^4$. So that's consistent.

Similarly, from equation $y^2 = 2 z^4$, so $y = \sqrt{2} z^2$, which is consistent.

So now, we can express all variables in terms of z. Let me do that.

Let me write:

x = √2 z

y = √2 z²

We also have the constraint xyz = 2. Let me substitute these into the constraint.

xyz = (√2 z) * (√2 z²) * z = Let me compute step by step.

First, √2 * √2 = 2

Then, z * z² * z = z^{1 + 2 + 1} = z^4

Therefore, xyz = 2 * z^4 = 2

Therefore:

2 z^4 = 2 => z^4 = 1 => z = 1 (since z is positive)

Therefore, z = 1.

Thus, x = √2 * 1 = √2

y = √2 * 1² = √2

So, x = √2, y = √2, z = 1

Let me check the constraint: xyz = √2 * √2 * 1 = (√2)^2 * 1 = 2 * 1 = 2. Correct.

Now, let's compute the value of the expression $x^4 + 4y^2 + 4z^4$.

Compute each term:

x^4 = (√2)^4 = (2^{1/2})^4 = 2^{2} = 4

4y^2 = 4*(√2)^2 = 4*2 = 8

4z^4 = 4*(1)^4 = 4*1 = 4

So total expression: 4 + 8 + 4 = 16.

Therefore, according to this, the minimum value is 16?

Wait, but let me check if this is indeed the minimum. Maybe there's a mistake here? Let me verify.

Alternatively, maybe I can test another set of values. Suppose I take x, y, z such that xyz = 2. For example, take x = 2, y = 1, z = 1. Then xyz = 2*1*1 = 2. Compute the expression: x^4 + 4y^2 + 4z^4 = 16 + 4 + 4 = 24, which is higher than 16. So 16 seems better.

Another test case: x = 1, y = 2, z = 1. Then xyz = 1*2*1 = 2. The expression is 1 + 4*(4) + 4 = 1 + 16 + 4 = 21, still higher.

Another test case: Let me try x = √2, y = √2, z = 1. Which is the critical point we found. So 4 + 8 + 4 = 16.

What if I take another critical point? Wait, how do I know that this is the only critical point? Well, since we used Lagrange multipliers and found this solution, and the function is convex? Not sure. But given that the test cases give higher values, maybe 16 is indeed the minimum. Let me check if there's a way to confirm using AM-GM.

Earlier, when I tried splitting the terms, perhaps I messed up the splitting. Let me try again. Let me think of splitting the expression into terms such that when applying AM-GM, the product relates to the constraint.

We have the expression x^4 + 4y^2 + 4z^4. Let me consider the terms as follows:

Let me think of x^4 as one term, 4y^2 as two terms of 2y^2 each, and 4z^4 as four terms of z^4 each. Wait, this is similar to before, but maybe I can use weighted AM-GM. Alternatively, think of it as:

Let me write the expression as x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4. That is, 1 + 2 + 4 terms. Wait, total of 1 + 2 + 4 = 7 terms. Then, applying AM-GM:

( x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4 ) / 7 ≥ ( x^4 * (2y^2)^2 * (z^4)^4 )^{1/7}

Wait, compute the product inside:

x^4 * (2y^2)^2 * (z^4)^4 = x^4 * 4y^4 * z^{16} = 4 x^4 y^4 z^{16}

But from the constraint, xyz = 2, so xyz = 2. Let me express x^4 y^4 z^{16} in terms of (xyz)^something.

Note that x^4 y^4 z^{16} = (xyz)^4 * z^{8} = (2)^4 * z^8 = 16 z^8

So the product is 4*16 z^8 = 64 z^8. Therefore:

AM-GM gives:

[Expression]/7 ≥ (64 z^8)^{1/7}

But this is still dependent on z, so not helpful. Therefore, this approach might not give a constant lower bound.

Alternatively, perhaps I need to split into different numbers of terms. Maybe using different coefficients.

Alternatively, think about the expression as x^4 + 4y^2 + 4z^4 and consider variables substitution. Let me set variables to make exponents match.

Let me denote a = x^2, b = y, c = z^2. Then, the original expression becomes:

a^2 + 4b^2 + 4c^2.

The constraint is xyz = 2. Since x = sqrt(a), y = b, z = sqrt(c), then the constraint is sqrt(a) * b * sqrt(c) = 2, which is sqrt(a c) * b = 2. So, b sqrt(a c) = 2.

So we need to minimize a^2 + 4b^2 + 4c^2 with b sqrt(a c) = 2.

This seems a bit complex, but maybe we can apply Cauchy-Schwarz or AM-GM here.

Alternatively, let me try substitution. From the constraint: b = 2 / sqrt(a c). Substitute into the expression:

Expression becomes:

a^2 + 4*(2 / sqrt(a c))^2 + 4c^2 = a^2 + 4*(4 / (a c)) + 4c^2 = a^2 + 16/(a c) + 4c^2.

Now, we have variables a and c. Maybe we can set t = a c, but not sure. Alternatively, set variables such that we can apply AM-GM.

Alternatively, think of a^2 + 4c^2 as two terms and 16/(a c) as another term. Maybe apply AM-GM on three terms? Let me try:

Let me consider the three terms: a^2, 4c^2, and 16/(a c). Wait, but they are not symmetric. However, maybe I can adjust them.

Alternatively, think of splitting 16/(a c) into multiple terms. Let me think of splitting the expression into more terms.

Let me try:

Let me think of the expression as a^2 + 4c^2 + 16/(a c). Maybe apply AM-GM on three terms. Let me check:

By AM-GM:

(a^2 + 4c^2 + 16/(a c))/3 ≥ cube root( a^2 * 4c^2 * 16/(a c) )

Compute the product inside:

a^2 * 4c^2 * 16/(a c) = 4 * 16 * a^2 c^2 / (a c) ) = 64 a c.

Therefore, the cube root of 64 a c = cube root(64) * cube root(a c) = 4 * cube root(a c).

So, the inequality becomes:

(a^2 + 4c^2 + 16/(a c))/3 ≥ 4 * cube root(a c)

But this seems messy, since we still have cube root(a c). However, if we can relate a c to something. From the constraint, earlier, we had that b = 2 / sqrt(a c). But in this substitution, I don't think that connects directly. Alternatively, perhaps there's another way.

Alternatively, set variables such that a c is fixed? Not sure. Maybe this approach isn't the best.

Alternatively, go back to the previous result from Lagrange multipliers. Since we found that the critical point gives the value 16, and testing other points gives higher values, maybe 16 is indeed the minimum.

But let me check if there is another way to apply AM-GM. Let me think again.

Original expression: x^4 + 4y^2 + 4z^4.

Let me think of the terms as follows: x^4, 2y^2, 2y^2, z^4, z^4, z^4, z^4. As before, 7 terms. Then, applying AM-GM:

( x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4 ) / 7 ≥ (x^4 * (2y^2)^2 * (z^4)^4 )^{1/7}

Which simplifies to:

[Expression]/7 ≥ [x^4 * 4y^4 * z^{16}]^{1/7} = [4x^4 y^4 z^{16}]^{1/7}

But since xyz = 2, then x^4 y^4 z^{16} = (xyz)^4 * z^8 = 2^4 * z^8 = 16 z^8.

Thus, the right side is [4 * 16 z^8]^{1/7} = [64 z^8]^{1/7} = 64^{1/7} * z^{8/7}

But since z is a variable, unless z is fixed, this won't help. However, if we can relate z to something else. From the critical point found earlier, z = 1. If we assume that the minimum occurs when z = 1, then maybe?

Alternatively, perhaps there's an error in splitting the terms. Maybe I need to split the expression into different numbers of terms. Let me try another approach.

Suppose I use AM-GM on x^4 and 4z^4. Let me consider x^4 and 4z^4 as two terms. Then, by AM-GM:

(x^4 + 4z^4)/2 ≥ sqrt(x^4 * 4z^4) = sqrt(4x^4 z^4) = 2x^2 z^2.

Therefore, x^4 + 4z^4 ≥ 4x^2 z^2.

Similarly, for the 4y^2 term, maybe pair with something. But we have 4y^2. If we consider 4y^2, maybe combine with other terms?

Alternatively, if I consider the entire expression:

x^4 + 4y^2 + 4z^4 ≥ 4x^2 z^2 + 4y^2.

But then, we can apply AM-GM on 4x^2 z^2 and 4y^2. Let me consider them as two terms:

(4x^2 z^2 + 4y^2)/2 ≥ sqrt(4x^2 z^2 * 4y^2) = sqrt(16x^2 y^2 z^2) = 4xyz.

Therefore, 4x^2 z^2 + 4y^2 ≥ 8xyz.

Therefore, the original expression:

x^4 + 4y^2 + 4z^4 ≥ 4x^2 z^2 + 4y^2 ≥ 8xyz.

Given that xyz = 2, so this would give:

Original expression ≥ 8 * 2 = 16.

Therefore, the minimum value is at least 16. And since we found a case where the expression equals 16, which satisfies the equality condition, that must be the minimum.

Wait, but how does the equality condition hold?

For AM-GM equality, all terms must be equal. So in the first step, when we applied AM-GM to x^4 and 4z^4, we have equality when x^4 = 4z^4 → x^2 = 2 z^2 → x = √2 z (since positive). Then, for the second AM-GM step, we applied to 4x^2 z^2 and 4y^2, so equality when 4x^2 z^2 = 4y^2 → x^2 z^2 = y^2 → y = x z. But from earlier, we had x = √2 z, so substituting:

y = x z = √2 z * z = √2 z^2. Which is exactly the relations found via Lagrange multipliers. Then, with xyz = 2, substituting x = √2 z and y = √2 z^2, we get:

xyz = √2 z * √2 z^2 * z = 2 z^4 = 2 → z^4 = 1 → z = 1. Therefore, all equality conditions are satisfied. Therefore, the minimum value is indeed 16. So this confirms the result from Lagrange multipliers.

Therefore, the minimum value of the expression is 16.

I think this is solid. The AM-GM approach gives the lower bound of 16, and we have a point where equality is achieved, so that must be the minimum. So the answer is 16.

**Final Answer**
\boxed{16}

To find the minimum value of the expression
$$
x^4 + 4y^2 + 4z^4
$$
under the constraint
$$
xyz = 2, \quad \text{where } x, y, z > 0,
$$
we consider the application of the **AM-GM Inequality** and the **Method of Lagrange Multipliers**.

---

### Step 1: Apply the AM-GM Inequality

We observe the expression and consider a modified form of the AM-GM inequality. Let us consider the following breakdown of the expression:

$$
x^4 + 4y^2 + 4z^4 = x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4
$$

This partitions the expression into **7 terms**:
- One term of $ x^4 $,
- Two terms of $ 2y^2 $, and
- Four terms of $ z^4 $.

Now, applying the AM-GM inequality to these 7 terms:

$$
\frac{x^4 + 2y^2 + 2y^2 + z^4 + z^4 + z^4 + z^4}{7} \geq \sqrt[7]{x^4 \cdot (2y^2)^2 \cdot (z^4)^4}
$$

Simplifying the right-hand side:

$$
\sqrt[7]{x^4 \cdot 4y^4 \cdot z^{16}} = \sqrt[7]{4x^4 y^4 z^{16}}
$$

But $ xyz = 2 $, so
$$
x^4 y^4 z^{16} = (xyz)^4 \cdot z^8 = 2^4 \cdot z^8 = 16z^8
$$

So the inequality becomes:
$$
\frac{x^4 + 4y^2 + 4z^4}{7} \geq \sqrt[7]{4 \cdot 16z^8} = \sqrt[7]{64z^8}
$$

Thus,
$$
x^4 + 4y^2 + 4z^4 \geq 7 \cdot \sqrt[7]{64z^8}
$$

But this result still depends on $ z $. To find a **constant** lower bound, we must ensure that the product inside the root is a constant. This suggests more symmetric reasoning is necessary.

---

### Step 2: Apply AM-GM on Two Pairs

Let us now try splitting the expression into two parts. Consider the inequality:
$$
x^4 + 4z^4 \geq 2\sqrt{x^4 \cdot 4z^4} = 2\sqrt{4x^4 z^4} = 4x^2 z^2
$$

So we have:
$$
x^4 + 4y^2 + 4z^4 \geq 4x^2 z^2 + 4y^2
$$

Now, apply AM-GM to the two terms $ 4x^2 z^2 $ and $ 4y^2 $:
$$
\frac{4x^2 z^2 + 4y^2}{2} \geq \sqrt{4x^2 z^2 \cdot 4y^2} = \sqrt{16x^2 y^2 z^2} = 4xyz
$$

Thus:
$$
4x^2 z^2 + 4y^2 \geq 8xyz
$$

Therefore:
$$
x^4 + 4y^2 + 4z^4 \geq 8xyz = 8 \cdot 2 = 16
$$

This gives the **lower bound** of **16**.

---

### Step 3: Check for Equality

To confirm that the **minimum value of 16** is achievable, we find values of $ x, y, z $ such that all AM-GM equalities are satisfied. That happens when:

- $ x^4 = 4z^4 $ → $ x^2 = 2z^2 $ → $ x = \sqrt{2}z $
- $ y^2 = x^2 z^2 = 2z^4 $ → $ y = \sqrt{2}z^2 $

So, substituting back into the constraint $ xyz = 2 $:
$$
x = \sqrt{2}z, \quad y = \sqrt{2}z^2 \Rightarrow xyz = \sqrt{2}z \cdot \sqrt{2}z^2 \cdot z = 2z^4 = 2 \Rightarrow z^4 = 1 \Rightarrow z = 1
$$

Hence:
- $ x = \sqrt{2} $
- $ y = \sqrt{2} $
- $ z = 1 $

Now, substitute these into the original expression:

$$
x^4 + 4y^2 + 4z^4 = (\sqrt{2})^4 + 4(\sqrt{2})^2 + 4(1)^4 = 4 + 8 + 4 = 16
$$

So, the **minimum value is indeed achieved** at $ x = \sqrt{2}, y = \sqrt{2}, z = 1 $, and the minimum value of the expression is:

$$
\boxed{16}
$$

Solution:
By AM-GM,
\begin{align*}
x^4 + 4y^2 + 4z^4 &= x^4 + 2y^2 + 2y^2 + 4z^4 \\
&\ge 4 \sqrt[4]{(x^4)(2y^2)(2y^2)(4z^4)} \\
&= 8xyz \\
&= 16.
\end{align*}Equality occurs when $x^4 = 2y^2 = 4z^2.$  Using the condition $xyz = 2,$ we can solve to get $x = y = \sqrt{2}$ and $z = 1,$ so the minimum value is $\boxed{16}.$


Explanation:
Okay, let's try to figure out what f(2015, 2) is. The function f is defined recursively, so I need to understand its behavior step by step. Let me start by looking at the definition again.

The function f takes two arguments: a non-negative integer i and a value j between 0 and 4. The function is defined recursively with three cases:

1. If i = 0, then f(0, j) = mod5(j + 1). So for i=0, it's straightforward. For example, f(0,0) would be mod5(0+1)=1, f(0,1)=2, etc., up to f(0,4)=mod5(4+1)=0.

2. If i ≥ 1 and j = 0, then f(i, 0) = f(i - 1, 1). So this is a recursive call where we decrease the first argument by 1 and set j to 1.

3. If i ≥ 1 and 1 ≤ j ≤ 4, then f(i, j) = f(i - 1, f(i, j - 1)). So this is another recursive call, but here we have to compute f(i, j - 1) first, then use that result as the second argument for f(i - 1, ...).

Hmm. This seems like a function that's building up some kind of recursive structure. Maybe similar to a function that iterates operations or something? Let me try calculating some small values of f(i, j) to see if I can find a pattern.

First, let's compute f(0, j) for all j from 0 to 4:

- f(0,0) = mod5(0 + 1) = 1
- f(0,1) = mod5(1 + 1) = 2
- f(0,2) = mod5(2 + 1) = 3
- f(0,3) = mod5(3 + 1) = 4
- f(0,4) = mod5(4 + 1) = 0

So for i=0, the values are just 1,2,3,4,0.

Now, let's compute f(1, j) for j=0 to 4.

Starting with j=0: f(1,0) = f(0,1) = 2 (from above).

Now j=1: f(1,1) = f(0, f(1,0)) = f(0, 2) = 3.

Similarly, j=2: f(1,2) = f(0, f(1,1)) = f(0, 3) = 4.

j=3: f(1,3) = f(0, f(1,2)) = f(0, 4) = 0.

j=4: f(1,4) = f(0, f(1,3)) = f(0, 0) = 1.

So f(1, j) is 2, 3, 4, 0, 1 for j=0 to 4.

Now let's compute f(2, j). Starting with j=0:

f(2,0) = f(1,1) = 3.

j=1: f(2,1) = f(1, f(2,0)) = f(1,3) = 0.

j=2: f(2,2) = f(1, f(2,1)) = f(1,0) = 2.

j=3: f(2,3) = f(1, f(2,2)) = f(1,2) = 4.

j=4: f(2,4) = f(1, f(2,3)) = f(1,4) = 1.

So f(2, j) is 3, 0, 2, 4, 1.

Moving on to f(3, j):

j=0: f(3,0) = f(2,1) = 0.

j=1: f(3,1) = f(2, f(3,0)) = f(2,0) = 3.

j=2: f(3,2) = f(2, f(3,1)) = f(2,3) = 4.

j=3: f(3,3) = f(2, f(3,2)) = f(2,4) = 1.

j=4: f(3,4) = f(2, f(3,3)) = f(2,1) = 0.

So f(3, j) is 0, 3, 4, 1, 0.

Now f(4, j):

j=0: f(4,0) = f(3,1) = 3.

j=1: f(4,1) = f(3, f(4,0)) = f(3,3) = 1.

j=2: f(4,2) = f(3, f(4,1)) = f(3,1) = 3.

j=3: f(4,3) = f(3, f(4,2)) = f(3,3) = 1.

j=4: f(4,4) = f(3, f(4,3)) = f(3,1) = 3.

So f(4, j) is 3, 1, 3, 1, 3.

Wait, that's interesting. Let me check that again.

For f(4,0): f(3,1)=3, which is correct.

f(4,1)=f(3, f(4,0))=f(3,3)=1, which matches.

f(4,2)=f(3, f(4,1))=f(3,1)=3.

f(4,3)=f(3, f(4,2))=f(3,3)=1.

f(4,4)=f(3, f(4,3))=f(3,1)=3. Yes, that's right.

So f(4, j) alternates between 3 and 1 starting from j=0? Wait, no, for j=0: 3; j=1:1; j=2:3; j=3:1; j=4:3. So it's 3,1,3,1,3.

Hmm. Let me compute f(5, j):

j=0: f(5,0) = f(4,1) = 1.

j=1: f(5,1) = f(4, f(5,0)) = f(4,1) = 1.

j=2: f(5,2) = f(4, f(5,1)) = f(4,1) = 1.

j=3: f(5,3) = f(4, f(5,2)) = f(4,1) = 1.

j=4: f(5,4) = f(4, f(5,3)) = f(4,1) = 1.

So f(5, j) is 1, 1, 1, 1, 1.

Wait, that's interesting! For i=5, all j's give 1? Let me check again:

For f(5,0) = f(4,1) = 1.

Then f(5,1) = f(4, f(5,0)) = f(4,1) = 1. Similarly, all subsequent ones will be f(4,1) which is 1. So yes, all f(5,j) = 1.

Wait, so if i=5, all j's result in 1? Let me check f(4,1)=1, which was from earlier. Yes, correct.

Now let's compute f(6, j):

j=0: f(6,0) = f(5,1) = 1.

j=1: f(6,1) = f(5, f(6,0)) = f(5,1) = 1.

Similarly, all j's for f(6, j) will be 1. So same as f(5, j). So starting from i=5, all higher i's will have f(i, j)=1 for all j?

Wait, let me check with f(6, j):

Yes, because when i ≥ 5, f(i,0) = f(i-1, 1). But if i-1 ≥5, then f(i-1,1) =1, so f(i,0)=1. Then for j=1, f(i,1)=f(i-1, f(i,0))=f(i-1,1)=1. Similarly, all subsequent j's. So once we reach i=5, all subsequent i's will have f(i,j)=1 for all j. Therefore, for i ≥5, f(i, j)=1.

Wait, but then the question is asking for f(2015, 2). Since 2015 is way larger than 5, then f(2015, 2) should be 1? But let me check if my reasoning is correct here.

Wait, let me check for i=5, j=0 to 4:

f(5, j) =1, yes. Then for i=6:

j=0: f(6,0) = f(5,1) =1.

j=1: f(6,1)=f(5, f(6,0))=f(5,1)=1.

j=2: f(6,2)=f(5, f(6,1))=f(5,1)=1.

Same for others. So yes, once we reach i=5, all subsequent i's will result in 1 for all j. Therefore, f(2015, 2) should be 1. But let me check with the previous steps again to make sure I didn't make a mistake.

Wait, but when I calculated f(4, j) as 3,1,3,1,3. Then f(5, j) is all 1's. Then f(6, j) is all 1's. So for any i ≥5, f(i, j) is 1? Let me check with i=5 and j=0 to 4, then i=6. Then yes, that seems to hold.

But wait, maybe there's a cycle or something. Let me check with i=4. For i=4, j=0: 3. Then for i=5, j=0: f(4,1)=1. Then for i=5, j=1: f(4, f(5,0))=f(4,1)=1. Then for i=5, j=2: f(4, f(5,1))=f(4,1)=1. Similarly for all others. So yes, once we reach i=5, everything becomes 1. Therefore, for any i ≥5, f(i, j) is 1. Hence, f(2015,2)=1.

Wait, but let me check for i=3. Earlier, f(3, j) was 0,3,4,1,0. Then i=4: f(4, j)=3,1,3,1,3. Then i=5: all 1's. So seems that from i=5 onwards, all values are 1. Therefore, f(2015, 2)=1?

But the problem says "for every positive integer n", but the function f is defined for i being a non-negative integer. So i=0,1,2,..., and j is from 0 to 4. So 2015 is a positive integer, so i=2015, j=2.

But according to my calculation, f(2015, 2)=1. But maybe I made a mistake?

Wait, let me check with i=5, j=0: f(5,0)=f(4,1)=1. Then f(5,1)=f(4, f(5,0))=f(4,1)=1. Then f(5,2)=f(4, f(5,1))=f(4,1)=1, and so on. So yes, for all j, f(5,j)=1.

Then f(6,0)=f(5,1)=1. Then same for all others. So for i ≥5, all f(i,j)=1.

Therefore, f(2015,2)=1. But let me check with smaller i's to make sure.

Wait, let me compute f(5,2). From above, f(5,2)=1. Let me check again:

f(5,0) = f(4,1) = 1.

f(5,1) = f(4, f(5,0)) = f(4,1) = 1.

f(5,2) = f(4, f(5,1)) = f(4,1) = 1. Yes. Then f(5,3) is f(4, f(5,2))=f(4,1)=1, and same for f(5,4).

So yes, f(5, j) is 1 for all j. Then f(6, j) would be same as f(5, j), since f(6,0)=f(5,1)=1, and so on.

Therefore, for any i ≥5, f(i, j)=1. Therefore, f(2015,2)=1.

But wait, the answer seems too straightforward. Let me check again with i=4. For i=4, j=0:3; j=1:1; j=2:3; j=3:1; j=4:3. Then i=5: all 1s. So maybe there's a cycle here?

But let me check for i=6, j=0: f(6,0)=f(5,1)=1. Then i=6, j=1: f(6,1)=f(5, f(6,0))=f(5,1)=1. Then i=6, j=2: f(6,2)=f(5, f(6,1))=f(5,1)=1. So yeah, same for all. So if i=5 is 1 for all, then i=6 is also 1 for all. So yeah, from i=5 onwards, it's 1. Therefore, f(2015, 2)=1. So is that the answer? But maybe I need to check more carefully.

Wait, maybe my assumption is wrong? Let me think about how the recursion works. Let me think of the function f(i, j) as something like a "hyperoperation" or a repeated function application? Maybe similar to the Ackermann function?

Alternatively, maybe there's a pattern in the values. Let me list out the values for different i and j:

For i=0:

[1, 2, 3, 4, 0]

i=1:

[2, 3, 4, 0, 1]

i=2:

[3, 0, 2, 4, 1]

i=3:

[0, 3, 4, 1, 0]

i=4:

[3, 1, 3, 1, 3]

i=5:

[1, 1, 1, 1, 1]

i=6:

[1, 1, 1, 1, 1]

And so on.

So yes, once i reaches 5, all values are 1. Therefore, for any i ≥ 5, f(i, j) is 1. Therefore, f(2015, 2) is 1. So the answer is 1? But let me check for i=5, j=0: f(5,0) = f(4,1) = 1. Then f(5,1) = f(4, f(5,0)) = f(4,1) =1. So yes, all the same.

Alternatively, maybe there's a cycle where after some i, it starts repeating. But according to the calculations, once it hits 1s, it stays at 1s.

But let me check i=5, j=0: f(5,0)=f(4,1)=1. Then for i=5, j=1: f(5,1)=f(4, f(5,0))=f(4,1)=1. Similarly, for j=2, same thing. So yeah, all 1s.

Alternatively, maybe my mistake was in calculating f(4, j). Let me check again:

For i=4:

j=0: f(4,0) = f(3,1) = 3.

j=1: f(4,1)=f(3, f(4,0))=f(3,3)=1. From earlier, f(3,3)=1.

j=2: f(4,2)=f(3, f(4,1))=f(3,1)=3.

j=3: f(4,3)=f(3, f(4,2))=f(3,3)=1.

j=4: f(4,4)=f(3, f(4,3))=f(3,1)=3.

Yes, so f(4, j) is [3,1,3,1,3]. So yes, correct. Then i=5:

j=0: f(5,0)=f(4,1)=1.

j=1: f(5,1)=f(4, f(5,0))=f(4,1)=1.

So yes.

Alternatively, maybe there is a different way to model this function? Let me think recursively.

The function f(i, j) seems to be building up some kind of iteration. For example, when i=0, it's just adding 1 mod5. For i=1, it's applying the function once. For i=2, it's applying it twice? Maybe?

Alternatively, think of f(i, j) as applying some operation j times? Wait, maybe similar to the concept of iterated functions?

Wait, let me think of f(i, j) as follows:

For each i, f(i, j) is defined in terms of f(i-1, ...) which suggests some kind of nested recursion.

Alternatively, maybe there is a relation to the concept of the "hyperoperator" or something else. But maybe we can think of this function as follows:

For i=0, f(0, j) = (j+1) mod5.

For i=1, f(1, j) = f(0, f(1, j-1)) with f(1,0)=f(0,1). So starting from j=0, it's f(0,1)=2, then f(1,1)=f(0,2)=3, etc. So for i=1, f(1, j) = (j + 2) mod5? Wait:

Wait f(1,0)=2, which is (0 + 2) mod5 = 2.

f(1,1)=3=(1 + 2) mod5=3.

f(1,2)=4=(2 + 2) mod5=4.

f(1,3)=0=(3 + 2) mod5=5 mod5=0.

f(1,4)=1=(4 + 2)=6 mod5=1. Yes, so f(1, j) = (j + 2) mod5.

Similarly, for i=2, let's see:

f(2, j) = f(1, f(2, j-1)), with f(2,0)=f(1,1)=3.

So f(2,0)=3.

f(2,1)=f(1,3)=0.

f(2,2)=f(1,0)=2.

f(2,3)=f(1,2)=4.

f(2,4)=f(1,4)=1.

So sequence is [3, 0, 2, 4, 1].

Is there a pattern here? Let's see:

If we think of i=2 as some kind of operation. Let me see:

If we think of f(i, j) as (j + something) mod5, but for i=2, it's not linear.

Alternatively, perhaps for each i, there's a cycle. For i=0, cycle length 5. For i=1, cycle length 5. For i=2, cycle length 5? Let me check:

For i=2, [3,0,2,4,1]. If we compute f(2, j+5), would it be the same? Let me check for j=0: f(2,0)=3. If we compute f(2,5), but wait, j is from 0 to 4. So maybe the cycle repeats every 5? Not sure.

But for i=3, the sequence was [0, 3, 4, 1, 0]. For i=4: [3,1,3,1,3]. For i=5: [1,1,1,1,1].

So maybe each i corresponds to some sort of iteration? Perhaps related to the number of times you apply operations?

Alternatively, perhaps we can think of f(i, j) as representing the result of applying some operation i times. But not sure.

Alternatively, maybe we can think of f(i, j) as follows: for each i, when you compute f(i, j), it's equivalent to doing something like j-th operation with i layers. But I'm not sure.

Alternatively, looking at the values:

i=0: [1, 2, 3, 4, 0]

i=1: [2, 3, 4, 0, 1]

i=2: [3, 0, 2, 4, 1]

i=3: [0, 3, 4, 1, 0]

i=4: [3, 1, 3, 1, 3]

i=5: [1, 1, 1, 1, 1]

i=6: [1, 1, 1, 1, 1]

So it seems that for i=0 to i=4, the values are different, but from i=5 onward, they are all 1.

Therefore, since 2015 is much larger than 5, then f(2015, 2) is 1. Therefore, the answer is 1?

Wait, but let me check with i=5 and i=6 again.

Suppose i=5, j=2: f(5,2)=1. Then for i=6, j=2: f(6,2)=f(5, f(6,1))=f(5,1)=1. Since f(6,1)=f(5, f(6,0))=f(5,1)=1. Therefore, yes, all subsequent i's will have f(i,j)=1 for any j.

Therefore, the conclusion is that for i ≥5, f(i, j)=1. Therefore, regardless of j, the value is 1.

Therefore, the answer should be 1. However, to be thorough, let me check for i=4, j=0: 3. Then for i=5, j=0: f(4,1)=1. Then for i=5, j=1: f(4, f(5,0))=f(4,1)=1. Then for i=6, j=0: f(5,1)=1. Similarly, for i=6, j=1: f(5, f(6,0))=f(5,1)=1. Therefore, it seems to hold.

Alternatively, maybe there's an error in the calculation for i=5? Let me recalculate f(5, j):

Starting with i=5, j=0:

f(5,0) = f(4,1) = 1 (from earlier: f(4,1)=1)

f(5,1) = f(4, f(5,0)) = f(4,1) =1

f(5,2) = f(4, f(5,1)) = f(4,1)=1

f(5,3) = f(4, f(5,2)) = f(4,1)=1

f(5,4) = f(4, f(5,3)) = f(4,1)=1

Yes, so all 1s. Therefore, i=5 onwards are 1s. Hence, for i=2015, which is way larger than 5, f(2015, 2) is 1.

But just to make sure, maybe I can think of it in terms of mathematical induction.

Assume that for some i ≥5, f(i, j)=1 for all j. Let's check if it holds for i+1.

For i+1, j=0: f(i+1, 0)=f(i, 1). Since for i ≥5, f(i,1)=1. Therefore, f(i+1, 0)=1.

For j≥1, f(i+1, j)=f(i, f(i+1, j-1)). Since f(i+1, j-1) for j-1 ≥0, if j-1=0, then f(i+1, 0)=1, which gives f(i, 1)=1. For j-1 ≥1, f(i+1, j-1) would be 1 as well. Therefore, f(i, 1)=1. Hence, for all j, f(i+1, j)=1. Hence, by induction, for all i ≥5, f(i, j)=1. Therefore, the conclusion is correct.

Thus, f(2015, 2)=1. Therefore, the answer is 1. But wait, let me check again with the initial problem statement to make sure I didn't misread anything.

The problem says "for every positive integer n", but the function is defined for i being a non-negative integer. So i=0,1,2,... and j is 0-4. The question is asking for f(2015, 2). Since 2015 is a positive integer (and indeed, a non-negative integer). Therefore, according to the above reasoning, the answer is 1.

But maybe there's a mistake in my calculation for i=5?

Wait, let me check with i=4, j=1: f(4,1)=1. Then i=5, j=0: f(5,0)=f(4,1)=1. Then for i=5, j=1: f(5,1)=f(4, f(5,0))=f(4,1)=1. Then for i=5, j=2: f(5,2)=f(4, f(5,1))=f(4,1)=1. Then same for others. So yes, it's 1. Therefore, it seems correct.

Therefore, I think the answer is 1. But let me check once more with another approach.

Alternative approach: Let's consider the function f(i, j) as a kind of iterated function. For example, if we think of f(i, j) as applying some function j times, but with i layers. But maybe not. Alternatively, since for i=0, it's a simple increment. For i=1, it's a bit more complex. But given that for i=5 onwards, all values are 1, which is a fixed point, because applying the function again won't change it.

In other words, if f(i, j)=1 for all j, then f(i+1, 0)=f(i,1)=1, and f(i+1, j)=f(i, f(i+1, j-1))=f(i,1)=1. So once we reach 1, it stays 1. Therefore, the function stabilizes at 1 for i ≥5.

Therefore, the answer should be 1. So I think this is correct.

But to be absolutely sure, let me check i=5, j=2 again. Since f(5,2)=f(4, f(5,1))=f(4,1)=1. And f(4,1)=1. Therefore, yes. Therefore, f(5,2)=1. Similarly for other j's.

Therefore, I think the answer is 1. Therefore, the answer should be \boxed{1}.

**Final Answer**
\boxed{1}

To determine the value of $ f(2015, 2) $, we need to understand the recursive structure and behavior of the function $ f(i, j) $, defined as:

$$
f(i,j) =
\begin{cases}
\text{mod}_5(j+1) & \text{if } i = 0 \text{ and } 0 \le j \le 4, \\
f(i-1, 1) & \text{if } i \ge 1 \text{ and } j = 0, \\
f(i-1, f(i,j-1)) & \text{if } i \ge 1 \text{ and } 1 \le j \le 4.
\end{cases}
$$

---

### **Base Case (i = 0)**

The base case is straightforward:
$$
f(0, j) = \text{mod}_5(j + 1) = (j+1) \bmod 5
$$
This gives:
- $ f(0, 0) = 1 $
- $ f(0, 1) = 2 $
- $ f(0, 2) = 3 $
- $ f(0, 3) = 4 $
- $ f(0, 4) = 0 $

---

### **Recursive Cases**

We proceed by computing the values of $ f(i, j) $ for $ i = 1 $ to $ i = 5 $, and observe patterns.

- **For i = 1**:
  $$
  f(1, j) = f(0, f(1, j-1)), \quad \text{with } f(1, 0) = f(0, 1) = 2
  $$
  Computing step-by-step:
  - $ f(1, 0) = 2 $
  - $ f(1, 1) = f(0, 2) = 3 $
  - $ f(1, 2) = f(0, 3) = 4 $
  - $ f(1, 3) = f(0, 4) = 0 $
  - $ f(1, 4) = f(0, 0) = 1 $

  So: $ f(1, j) = [2, 3, 4, 0, 1] $

- **For i = 2**:
  Using $ f(2, 0) = f(1, 1) = 3 $, and recursively computing:
  - $ f(2, 0) = 3 $
  - $ f(2, 1) = f(1, 3) = 0 $
  - $ f(2, 2) = f(1, 0) = 2 $
  - $ f(2, 3) = f(1, 2) = 4 $
  - $ f(2, 4) = f(1, 4) = 1 $

  So: $ f(2, j) = [3, 0, 2, 4, 1] $

- **For i = 3**:
  $ f(3, 0) = f(2, 1) = 0 $
  - $ f(3, 1) = f(2, 0) = 3 $
  - $ f(3, 2) = f(2, 3) = 4 $
  - $ f(3, 3) = f(2, 4) = 1 $
  - $ f(3, 4) = f(2, 1) = 0 $

  So: $ f(3, j) = [0, 3, 4, 1, 0] $

- **For i = 4**:
  - $ f(4, 0) = f(3, 1) = 3 $
  - $ f(4, 1) = f(3, 3) = 1 $
  - $ f(4, 2) = f(3, 1) = 3 $
  - $ f(4, 3) = f(3, 3) = 1 $
  - $ f(4, 4) = f(3, 1) = 3 $

  So: $ f(4, j) = [3, 1, 3, 1, 3] $

- **For i = 5**:
  - $ f(5, 0) = f(4, 1) = 1 $
  - Then for all $ j \ge 1 $, the function recursively evaluates:
    $$
    f(i, j) = f(i-1, f(i, j-1)) = f(i-1, 1)
    $$
    Since $ f(4, 1) = 1 $, and recursively everything depends on $ f(i, 1) = 1 $, we get:
    $$
    f(5, j) = 1 \quad \text{for all } j
    $$

---

### **Inductive Argument**

Observe that once all of $ f(i, j) = 1 $ for all $ j $, then for any higher $ i + 1 $, the values will also be 1:

- $ f(i+1, 0) = f(i, 1) = 1 $
- $ f(i+1, j) = f(i, f(i+1, j-1)) = f(i, 1) = 1 $

Therefore, all values $ f(i, j) = 1 $ for all $ i \ge 5 $ and $ j \in \{0,1,2,3,4\} $

---

### **Final Computation**

Since $ 2015 \ge 5 $, and we've established that for all $ i \ge 5 $, $ f(i, j) = 1 $ for all $ j $, it follows that:

$$
f(2015, 2) = \boxed{1}
$$

[Assistant]
Explanation:
Okay, let's try to figure out what f(2015, 2) is. The function f is defined recursively, so I need to understand its behavior step by step. Let me start by looking at the definition again.

The function f takes two arguments: a non-negative integer i and a value j between 0 and 4. The function is defined recursively with three cases:

1. If i = 0, then f(0, j) = mod5(j + 1). So for i=0, it's straightforward. For example, f(0,0) would be mod5(0+1)=1, f(0,1)=2, etc., up to f(0,4)=mod5(4+1)=0.

2. If i ≥ 1 and j = 0, then f(i, 0) = f(i - 1, 1). So this is a recursive call where we decrease the first argument by 1 and set j to 1.

3. If i ≥ 1 and 1 ≤ j ≤ 4, then f(i, j) = f(i - 1, f(i, j - 1)). So this is another recursive call, but here we have to compute f(i, j - 1) first, then use that result as the second argument for f(i - 1, ...).

Hmm. This seems like a function that's building up some kind of recursive structure. Maybe similar to a function that iterates operations or something? Let me try calculating some small values of f(i, j) to see if I can find a pattern.

First, let's compute f(0, j) for all j from 0 to 4:

- f(0,0) = mod5(0 + 1) = 1
- f(0,1) = mod5(1 + 1) = 2
- f(0,2) = mod5(2 + 1) = 3
- f(0,3) = mod5(3 + 1) = 4
- f(0,4) = mod5(4 + 1) = 0

So for i=0, the values are just 1,2,3,4,0.

Now, let's compute f(1, j) for j=0 to 4.

Starting with j=0: f(1,0) = f(0,1) = 2 (from above).

Now j=1: f(1,1) = f(0, f(1,0)) = f(0, 2) = 3.

Similarly, j=2: f(1,2) = f(0, f(1,1)) = f(0, 3) = 4.

j=3: f(1,3) = f(0, f(1,2)) = f(0, 4) = 0.

j=4: f(1,4) = f(0, f(1,3)) = f(0, 0) = 1.

So f(1, j) is 2, 3, 4, 0, 1 for j=0 to 4.

Now let's compute f(2, j). Starting with j=0:

f(2,0) = f(1,1) = 3.

j=1: f(2,1) = f(1, f(2,0)) = f(1,3) = 0.

j=2: f(2,2) = f(1, f(2,1)) = f(1,0) = 2.

j=3: f(2,3) = f(1, f(2,2)) = f(1,2) = 4.

j=4: f(2,4) = f(1, f(2,3)) = f(1,4) = 1.

So f(2, j) is 3, 0, 2, 4, 1.

Moving on to f(3, j):

j=0: f(3,0) = f(2,1) = 0.

j=1: f(3,1) = f(2, f(3,0)) = f(2,0) = 3.

j=2: f(3,2) = f(2, f(3,1)) = f(2,3) = 4.

j=3: f(3,3) = f(2, f(3,2)) = f(2,4) = 1.

j=4: f(3,4) = f(2, f(3,3)) = f(2,1) = 0.

So f(3, j) is 0, 3, 4, 1, 0.

Now f(4, j):

j=0: f(4,0) = f(3,1) = 3.

j=1: f(4,1) = f(3, f(4,0)) = f(3,3) = 1.

j=2: f(4,2) = f(3, f(4,1)) = f(3,1) = 3.

j=3: f(4,3) = f(3, f(4,2)) = f(3,3) = 1.

j=4: f(4,4) = f(3, f(4,3)) = f(3,1) = 3.

So f(4, j) is 3, 1, 3, 1, 3.

Wait, that's interesting. Let me check that again.

For f(4,0): f(3,1)=3, which is correct.

f(4,1)=f(3, f(4,0))=f(3,3)=1, which matches.

f(4,2)=f(3, f(4,1))=f(3,1)=3.

f(4,3)=f(3, f(4,2))=f(3,3)=1.

f(4,4)=f(3, f(4,3))=f(3,1)=3. Yes, that's right.

So f(4, j) alternates between 3 and 1 starting from j=0? Wait, no, for j=0: 3; j=1:1; j=2:3; j=3:1; j=4:3. So it's 3,1,3,1,3.

Hmm. Let me compute f(5, j):

j=0: f(5,0) = f(4,1) = 1.

j=1: f(5,1) = f(4, f(5,0)) = f(4,1) = 1.

j=2: f(5,2) = f(4, f(5,1)) = f(4,1) = 1.

j=3: f(5,3) = f(4, f(5,2)) = f(4,1) = 1.

j=4: f(5,4) = f(4, f(5,3)) = f(4,1) = 1.

So f(5, j) is 1, 1, 1, 1, 1.

Wait, that's interesting! For i=5, all j's give 1? Let me check again:

For f(5,0) = f(4,1) = 1.

Then f(5,1) = f(4, f(5,0)) = f(4,1) = 1. Similarly, all subsequent ones will be f(4,1) which is 1. So yes, all f(5,j) = 1.

Wait, so if i=5, all j's result in 1? Let me check f(4,1)=1, which was from earlier. Yes, correct.

Now let's compute f(6, j):

j=0: f(6,0) = f(5,1) = 1.

j=1: f(6,1) = f(5, f(6,0)) = f(5,1) = 1.

Similarly, all j's for f(6, j) will be 1. So same as f(5, j). So starting from i=5, all higher i's will have f(i, j)=1 for all j?

Wait, let me check with f(6, j):

Yes, because when i ≥ 5, f(i,0) = f(i-1, 1). But if i-1 ≥5, then f(i-1,1) =1, so f(i,0)=1. Then for j=1, f(i,1)=f(i-1, f(i,0))=f(i-1,1)=1. Similarly, all subsequent j's. So once we reach i=5, all subsequent i's will have f(i,j)=1 for all j. Therefore, for i ≥5, f(i, j)=1.

Wait, but then the question is asking for f(2015, 2). Since 2015 is way larger than 5, then f(2015, 2) should be 1? But let me check if my reasoning is correct here.

Wait, let me check for i=5, j=0 to 4:

f(5, j) =1, yes. Then for i=6:

j=0: f(6,0) = f(5,1) =1.

j=1: f(6,1)=f(5, f(6,0))=f(5,1)=1.

j=2: f(6,2)=f(5, f(6,1))=f(5,1)=1.

Same for others. So yes, once we reach i=5, all subsequent i's will result in 1 for all j. Therefore, f(2015, 2) should be 1. But let me check with the previous steps again to make sure I didn't make a mistake.

Wait, but when I calculated f(4, j) as 3,1,3,1,3. Then f(5, j) is all 1's. Then f(6, j) is all 1's. So for any i ≥5, f(i, j) is 1? Let me check with i=5 and j=0 to 4, then i=6. Then yes, that seems to hold.

But wait, maybe there's a cycle or something. Let me check with i=4. For i=4, j=0: 3. Then for i=5, j=0: f(4,1)=1. Then for i=5, j=1: f(4, f(5,0))=f(4,1)=1. Then for i=5, j=2: f(4, f(5,1))=f(4,1)=1. Similarly for all others. So yes, once we reach i=5, everything becomes 1. Therefore, for any i ≥5, f(i, j) is 1. Hence, f(2015,2)=1.

Wait, but let me check for i=3. Earlier, f(3, j) was 0,3,4,1,0. Then i=4: f(4, j)=3,1,3,1,3. Then i=5: all 1's. So seems that from i=5 onwards, all values are 1. Therefore, f(2015, 2)=1?

But the problem says "for every positive integer n", but the function f is defined for i being a non-negative integer. So i=0,1,2,..., and j is from 0 to 4. So 2015 is a positive integer, so i=2015, j=2.

But according to my calculation, f(2015, 2)=1. But maybe I made a mistake?

Wait, let me check with i=5, j=0: f(5,0)=f(4,1)=1. Then f(5,1)=f(4, f(5,0))=f(4,1)=1. Then f(5,2)=f(4, f(5,1))=f(4,1)=1, and so on. So yes, for all j, f(5,j)=1.

Then f(6,0)=f(5,1)=1. Then same for all others. So for i ≥5, all f(i,j)=1.

Therefore, f(2015,2)=1. But let me check with smaller i's to make sure.

Wait, let me compute f(5,2). From above, f(5,2)=1. Let me check again:

f(5,0) = f(4,1) = 1.

f(5,1) = f(4, f(5,0)) = f(4,1) = 1.

f(5,2) = f(4, f(5,1)) = f(4,1) = 1. Yes. Then f(5,3) is f(4, f(5,2))=f(4,1)=1, and same for f(5,4).

So yes, f(5, j) is 1 for all j. Then f(6, j) would be same as f(5, j), since f(6,0)=f(5,1)=1, and so on.

Therefore, for any i ≥5, f(i, j)=1. Therefore, f(2015,2)=1.

But wait, the answer seems too straightforward. Let me check again with i=4. For i=4, j=0:3; j=1:1; j=2:3; j=3:1; j=4:3. Then i=5: all 1s. So maybe there's a cycle here?

But let me check for i=6, j=0: f(6,0)=f(5,1)=1. Then i=6, j=1: f(6,1)=f(5, f(6,0))=f(5,1)=1. Then i=6, j=2: f(6,2)=f(5, f(6,1))=f(5,1)=1. So yeah, same for all. So if i=5 is 1 for all, then i=6 is also 1 for all. So yeah, from i=5 onwards, it's 1. Therefore, f(2015, 2)=1. So is that the answer? But maybe I need to check more carefully.

Wait, maybe my assumption is wrong? Let me think about how the recursion works. Let me think of the function f(i, j) as something like a "hyperoperation" or a repeated function application? Maybe similar to the Ackermann function?

Alternatively, maybe there's a pattern in the values. Let me list out the values for different i and j:

For i=0:

[1, 2, 3, 4, 0]

i=1:

[2, 3, 4, 0, 1]

i=2:

[3, 0, 2, 4, 1]

i=3:

[0, 3, 4, 1, 0]

i=4:

[3, 1, 3, 1, 3]

i=5:

[1, 1, 1, 1, 1]

i=6:

[1, 1, 1, 1, 1]

And so on.

So yes, once i reaches 5, all values are 1. Therefore, for any i ≥ 5, f(i, j) is 1. Therefore, f(2015, 2) is 1. So the answer is 1? But let me check for i=5, j=0: f(5,0) = f(4,1) = 1. Then f(5,1) = f(4, f(5,0)) = f(4,1) =1. So yes, all the same.

Alternatively, maybe there's a cycle where after some i, it starts repeating. But according to the calculations, once it hits 1s, it stays at 1s.

But let me check i=5, j=0: f(5,0)=f(4,1)=1. Then for i=5, j=1: f(5,1)=f(4, f(5,0))=f(4,1)=1. Similarly, for j=2, same thing. So yeah, all 1s.

Alternatively, maybe my mistake was in calculating f(4, j). Let me check again:

For i=4:

j=0: f(4,0) = f(3,1) = 3.

j=1: f(4,1)=f(3, f(4,0))=f(3,3)=1. From earlier, f(3,3)=1.

j=2: f(4,2)=f(3, f(4,1))=f(3,1)=3.

j=3: f(4,3)=f(3, f(4,2))=f(3,3)=1.

j=4: f(4,4)=f(3, f(4,3))=f(3,1)=3.

Yes, so f(4, j) is [3,1,3,1,3]. So yes, correct. Then i=5:

j=0: f(5,0)=f(4,1)=1.

j=1: f(5,1)=f(4, f(5,0))=f(4,1)=1.

So yes.

Alternatively, maybe there is a different way to model this function? Let me think recursively.

The function f(i, j) seems to be building up some kind of iteration. For example, when i=0, it's just adding 1 mod5. For i=1, it's applying the function once. For i=2, it's applying it twice? Maybe?

Alternatively, think of f(i, j) as applying some operation j times? Wait, maybe similar to the concept of iterated functions?

Wait, let me think of f(i, j) as follows:

For each i, f(i, j) is defined in terms of f(i-1, ...) which suggests some kind of nested recursion.

Alternatively, maybe there is a relation to the concept of the "hyperoperator" or something else. But maybe we can think of this function as follows:

For i=0, f(0, j) = (j+1) mod5.

For i=1, f(1, j) = f(0, f(1, j-1)) with f(1,0)=f(0,1). So starting from j=0, it's f(0,1)=2, then f(1,1)=f(0,2)=3, etc. So for i=1, f(1, j) = (j + 2) mod5? Wait:

Wait f(1,0)=2, which is (0 + 2) mod5 = 2.

f(1,1)=3=(1 + 2) mod5=3.

f(1,2)=4=(2 + 2) mod5=4.

f(1,3)=0=(3 + 2) mod5=5 mod5=0.

f(1,4)=1=(4 + 2)=6 mod5=1. Yes, so f(1, j) = (j + 2) mod5.

Similarly, for i=2, let's see:

f(2, j) = f(1, f(2, j-1)), with f(2,0)=f(1,1)=3.

So f(2,0)=3.

f(2,1)=f(1,3)=0.

f(2,2)=f(1,0)=2.

f(2,3)=f(1,2)=4.

f(2,4)=f(1,4)=1.

So sequence is [3, 0, 2, 4, 1].

Is there a pattern here? Let's see:

If we think of i=2 as some kind of operation. Let me see:

If we think of f(i, j) as (j + something) mod5, but for i=2, it's not linear.

Alternatively, perhaps for each i, there's a cycle. For i=0, cycle length 5. For i=1, cycle length 5. For i=2, cycle length 5? Let me check:

For i=2, [3,0,2,4,1]. If we compute f(2, j+5), would it be the same? Let me check for j=0: f(2,0)=3. If we compute f(2,5), but wait, j is from 0 to 4. So maybe the cycle repeats every 5? Not sure.

But for i=3, the sequence was [0, 3, 4, 1, 0]. For i=4: [3,1,3,1,3]. For i=5: [1,1,1,1,1].

So maybe each i corresponds to some sort of iteration? Perhaps related to the number of times you apply operations?

Alternatively, perhaps we can think of f(i, j) as representing the result of applying some operation i times. But not sure.

Alternatively, maybe we can think of f(i, j) as follows: for each i, when you compute f(i, j), it's equivalent to doing something like j-th operation with i layers. But I'm not sure.

Alternatively, looking at the values:

i=0: [1, 2, 3, 4, 0]

i=1: [2, 3, 4, 0, 1]

i=2: [3, 0, 2, 4, 1]

i=3: [0, 3, 4, 1, 0]

i=4: [3, 1, 3, 1, 3]

i=5: [1, 1, 1, 1, 1]

i=6: [1, 1, 1, 1, 1]

So it seems that for i=0 to i=4, the values are different, but from i=5 onward, they are all 1.

Therefore, since 2015 is much larger than 5, then f(2015, 2) is 1. Therefore, the answer is 1?

Wait, but let me check with i=5 and i=6 again.

Suppose i=5, j=2: f(5,2)=1. Then for i=6, j=2: f(6,2)=f(5, f(6,1))=f(5,1)=1. Since f(6,1)=f(5, f(6,0))=f(5,1)=1. Therefore, yes, all subsequent i's will have f(i,j)=1 for any j.

Therefore, the conclusion is that for i ≥5, f(i, j)=1. Therefore, regardless of j, the value is 1.

Therefore, the answer should be 1. However, to be thorough, let me check for i=4, j=0: 3. Then for i=5, j=0: f(4,1)=1. Then for i=5, j=1: f(4, f(5,0))=f(4,1)=1. Then for i=6, j=0: f(5,1)=1. Similarly, for i=6, j=1: f(5, f(6,0))=f(5,1)=1. Therefore, it seems to hold.

Alternatively, maybe there's an error in the calculation for i=5? Let me recalculate f(5, j):

Starting with i=5, j=0:

f(5,0) = f(4,1) = 1 (from earlier: f(4,1)=1)

f(5,1) = f(4, f(5,0)) = f(4,1) =1

f(5,2) = f(4, f(5,1)) = f(4,1)=1

f(5,3) = f(4, f(5,2)) = f(4,1)=1

f(5,4) = f(4, f(5,3)) = f(4,1)=1

Yes, so all 1s. Therefore, i=5 onwards are 1s. Hence, for i=2015, which is way larger than 5, f(2015, 2) is 1.

But just to make sure, maybe I can think of it in terms of mathematical induction.

Assume that for some i ≥5, f(i, j)=1 for all j. Let's check if it holds for i+1.

For i+1, j=0: f(i+1, 0)=f(i, 1). Since for i ≥5, f(i,1)=1. Therefore, f(i+1, 0)=1.

For j≥1, f(i+1, j)=f(i, f(i+1, j-1)). Since f(i+1, j-1) for j-1 ≥0, if j-1=0, then f(i+1, 0)=1, which gives f(i, 1)=1. For j-1 ≥1, f(i+1, j-1) would be 1 as well. Therefore, f(i, 1)=1. Hence, for all j, f(i+1, j)=1. Hence, by induction, for all i ≥5, f(i, j)=1. Therefore, the conclusion is correct.

Thus, f(2015, 2)=1. Therefore, the answer is 1. But wait, let me check again with the initial problem statement to make sure I didn't misread anything.

The problem says "for every positive integer n", but the function is defined for i being a non-negative integer. So i=0,1,2,... and j is 0-4. The question is asking for f(2015, 2). Since 2015 is a positive integer (and indeed, a non-negative integer). Therefore, according to the above reasoning, the answer is 1.

But maybe there's a mistake in my calculation for i=5?

Wait, let me check with i=4, j=1: f(4,1)=1. Then i=5, j=0: f(5,0)=f(4,1)=1. Then for i=5, j=1: f(5,1)=f(4, f(5,0))=f(4,1)=1. Then for i=5, j=2: f(5,2)=f(4, f(5,1))=f(4,1)=1. Then same for others. So yes, it's 1. Therefore, it seems correct.

Therefore, I think the answer is 1. But let me check once more with another approach.

Alternative approach: Let's consider the function f(i, j) as a kind of iterated function. For example, if we think of f(i, j) as applying some function j times, but with i layers. But maybe not. Alternatively, since for i=0, it's a simple increment. For i=1, it's a bit more complex. But given that for i=5 onwards, all values are 1, which is a fixed point, because applying the function again won't change it.

In other words, if f(i, j)=1 for all j, then f(i+1, 0)=f(i,1)=1, and f(i+1, j)=f(i, f(i+1, j-1))=f(i,1)=1. So once we reach 1, it stays 1. Therefore, the function stabilizes at 1 for i ≥5.

Therefore, the answer should be 1. So I think this is correct.

But to be absolutely sure, let me check i=5, j=2 again. Since f(5,2)=f(4, f(5,1))=f(4,1)=1. And f(4,1)=1. Therefore, yes. Therefore, f(5,2)=1. Similarly for other j's.

Therefore, I think the answer is 1. Therefore, the answer should be \boxed{1}.

**Final Answer**
\boxed{1}

To determine the value of $ f(2015, 2) $, we need to understand the recursive structure and behavior of the function $ f(i, j) $, defined as:

$$
f(i,j) =
\begin{cases}
\text{mod}_5(j+1) & \text{if } i = 0 \text{ and } 0 \le j \le 4, \\
f(i-1, 1) & \text{if } i \ge 1 \text{ and } j = 0, \\
f(i-1, f(i,j-1)) & \text{if } i \ge 1 \text{ and } 1 \le j \le 4.
\end{cases}
$$

---

### **Base Case (i = 0)**

The base case is straightforward:
$$
f(0, j) = \text{mod}_5(j + 1) = (j+1) \bmod 5
$$
This gives:
- $ f(0, 0) = 1 $
- $ f(0, 1) = 2 $
- $ f(0, 2) = 3 $
- $ f(0, 3) = 4 $
- $ f(0, 4) = 0 $

---

### **Recursive Cases**

We proceed by computing the values of $ f(i, j) $ for $ i = 1 $ to $ i = 5 $, and observe patterns.

- **For i = 1**:
  $$
  f(1, j) = f(0, f(1, j-1)), \quad \text{with } f(1, 0) = f(0, 1) = 2
  $$
  Computing step-by-step:
  - $ f(1, 0) = 2 $
  - $ f(1, 1) = f(0, 2) = 3 $
  - $ f(1, 2) = f(0, 3) = 4 $
  - $ f(1, 3) = f(0, 4) = 0 $
  - $ f(1, 4) = f(0, 0) = 1 $

  So: $ f(1, j) = [2, 3, 4, 0, 1] $

- **For i = 2**:
  Using $ f(2, 0) = f(1, 1) = 3 $, and recursively computing:
  - $ f(2, 0) = 3 $
  - $ f(2, 1) = f(1, 3) = 0 $
  - $ f(2, 2) = f(1, 0) = 2 $
  - $ f(2, 3) = f(1, 2) = 4 $
  - $ f(2, 4) = f(1, 4) = 1 $

  So: $ f(2, j) = [3, 0, 2, 4, 1] $

- **For i = 3**:
  $ f(3, 0) = f(2, 1) = 0 $
  - $ f(3, 1) = f(2, 0) = 3 $
  - $ f(3, 2) = f(2, 3) = 4 $
  - $ f(3, 3) = f(2, 4) = 1 $
  - $ f(3, 4) = f(2, 1) = 0 $

  So: $ f(3, j) = [0, 3, 4, 1, 0] $

- **For i = 4**:
  - $ f(4, 0) = f(3, 1) = 3 $
  - $ f(4, 1) = f(3, 3) = 1 $
  - $ f(4, 2) = f(3, 1) = 3 $
  - $ f(4, 3) = f(3, 3) = 1 $
  - $ f(4, 4) = f(3, 1) = 3 $

  So: $ f(4, j) = [3, 1, 3, 1, 3] $

- **For i = 5**:
  - $ f(5, 0) = f(4, 1) = 1 $
  - Then for all $ j \ge 1 $, the function recursively evaluates:
    $$
    f(i, j) = f(i-1, f(i, j-1)) = f(i-1, 1)
    $$
    Since $ f(4, 1) = 1 $, and recursively everything depends on $ f(i, 1) = 1 $, we get:
    $$
    f(5, j) = 1 \quad \text{for all } j
    $$

---

### **Inductive Argument**

Observe that once all of $ f(i, j) = 1 $ for all $ j $, then for any higher $ i + 1 $, the values will also be 1:

- $ f(i+1, 0) = f(i, 1) = 1 $
- $ f(i+1, j) = f(i, f(i+1, j-1)) = f(i, 1) = 1 $

Therefore, all values $ f(i, j) = 1 $ for all $ i \ge 5 $ and $ j \in \{0,1,2,3,4\} $

---

### **Final Computation**

Since $ 2015 \ge 5 $, and we've established that for all $ i \ge 5 $, $ f(i, j) = 1 $ for all $ j $, it follows that:

$$
f(2015, 2) = \boxed{1}
$$

Solution:
We construct a table for the values $f(i,j)$:
\[
\begin{array}{c|ccccc}
i \backslash j & 0 & 1 & 2 & 3 & 4 \\ \hline
0 & 1 & 2 & 3 & 4 & 0 \\
1 & 2 & 3 & 4 & 0 & 1 \\
2 & 3 & 0 & 2 & 4 & 1 \\
3 & 0 & 3 & 4 & 1 & 0 \\
4 & 3 & 1 & 3 & 1 & 3 \\
5 & 1 & 1 & 1 & 1 & 1 \\
6 & 1 & 1 & 1 & 1 & 1
\end{array}
\]It follows that $f(i,2) = \boxed{1}$ for all $i \ge 5.$