Powers equation with 3 variables

Since $x ^ {y} + y ^ {z} = z ^ {x}$ therefore $y ^ {z} < z ^ {x}$ therefore $y ^ {z} < z ^ {y}$

Raising both sides to the power $\frac{1}{yz}$ we get:

$y ^ \frac{1}{y} < z ^ \frac{1}{z}$

But since

$3 ^ \frac{1}{3} > 4 ^ \frac{1}{4} > 5 ^ \frac{1}{5} ...$

It follows that y < 3.

We have now 0 < x <= y < 3. This will give us these possibilities for (x, y) = (1, 1) or (1, 2) or (2, 2)

Let’s consider each case separately.

Case # 1.
If (x, y) = (1, 1). This will give us: $1 ^ {1} + 1 ^ {z} = z ^ {1}.$ This means that z = 2. The first possible solution is : x = 1, y = 1 and z = 2. Therefore v = 1 + 1 + 2 = 4.

Case # 2.
If (x, y) = (1, 2). In this case z will be greater than or equal 2. This will give us: $1 ^ {2} + 2 ^ {z} = z ^ {1}.$ This means that z > $2 ^ {z}$ which is impossible for any value of z. This case should be rejected.

Case # 3.
If (x, y) = (2, 2). Again in this case z will be greater than or equal 2. This will give us: $2 ^ {2} + 2 ^ {z} = z ^ {2}.$ This means that $z ^ {2} > 2 ^ {z}$ which is impossible for any value of z except when 2 < z < 4. This will leave us with z = 3. Substituting x = 2, y = 2 and z = 3 we find that it does not satisfy the equation. Therefore, this case should be rejected.

The only answer for v = 4.

$Answer = \boxed{4}$