We prove that all primitive Pythagorean triples $(a, b, c)$ with $a$ odd are given by $a = m^2 - n^2, b = 2mn, c = m^2 + n^2$ where $m > n > 0$, $\gcd(m, n) = 1$, and $m \not\equiv n \pmod{2}$.

Step 1: Verify the parametrization gives Pythagorean triples.

$(m^2 - n^2)^2 + (2mn)^2 = m^4 - 2m^2n^2 + n^4 + 4m^2n^2 = m^4 + 2m^2n^2 + n^4 = (m^2 + n^2)^2$ ✓

Step 2: Show every primitive triple arises this way.

Let $(a, b, c)$ be a primitive Pythagorean triple with $a$ odd. Then $b$ is even (if both $a, b$ were odd, $a^2 + b^2 \equiv 2 \pmod{4}$, but squares are $0$ or $1 \bmod 4$, impossible for $c^2$).

From $a^2 = c^2 - b^2 = (c-b)(c+b)$, write $b = 2b'$. Then: $a^2 + 4b'^2 = c^2 \implies a^2 = (c - 2b')(c + 2b')$

Since $\gcd(a, b, c) = 1$ and $a$ is odd, both $c - 2b'$ and $c + 2b'$ are odd.

Claim: $\gcd(c - 2b', c + 2b') = 1$.

Proof: Any common divisor $d$ divides their sum $2c$ and difference $4b'$. Since $\gcd(b, c) = 1$ and $d$ is odd, we have $d = 1$.

Therefore, $c - 2b'$ and $c + 2b'$ are both perfect squares: $c - 2b' = n^2, c + 2b' = m^2$ with $m > n > 0$.

Solving: $c = \frac{m^2 + n^2}{2}, b' = \frac{m^2 - n^2}{4}$.

Wait, let me reconsider. From $b^2 = c^2 - a^2$:

$b^2 = (c - a)(c + a)$

Since $\gcd(a, c) = 1$ (primitive), we have $\gcd(c - a, c + a) \leq 2$.

Since $c - a$ and $c + a$ have the same parity and their product is $b^2$ (even squared), both must be even. Write $c - a = 2u, c + a = 2v$ with $\gcd(u, v) = 1$.

Then $b^2 = 4uv$, so $b = 2\sqrt{uv}$. For $b$ to be an integer, $uv$ must be a perfect square. Since $\gcd(u, v) = 1$, both $u$ and $v$ are perfect squares: $u = n^2, v = m^2$.

Therefore: $c = u + v = n^2 + m^2, a = v - u = m^2 - n^2, b = 2\sqrt{uv} = 2mn$.

We prove no positive integers satisfy $x^4 + y^4 = z^2$ using Fermat's method of infinite descent.

Suppose $(x, y, z)$ is a solution with $x, y, z > 0$ and minimal $z$. We may assume $\gcd(x, y) = 1$ (otherwise divide by $\gcd(x, y)^4$).

Then $(x^2, y^2, z)$ is a Pythagorean triple. By the parametrization: $x^2 = m^2 - n^2, \quad y^2 = 2mn, \quad z = m^2 + n^2$ for some $m > n > 0$ with $\gcd(m, n) = 1$ and $m \not\equiv n \pmod{2}$.

From $y^2 = 2mn$: since $\gcd(m, n) = 1$ and their product (times $2$) is a square, we have two cases:

Case 1: $m = 2s^2, n = t^2$ with $\gcd(s, t) = 1$.

Then $x^2 = m^2 - n^2 = 4s^4 - t^4$. This gives $(t^2, 2s^2, x)$ as a Pythagorean triple: $t^4 + 4s^4 = x^2$

Applying parametrization again leads to a smaller solution.

Case 2: Similar analysis.

In both cases, we construct a Pythagorean triple $(a^2, b^2, c)$ with $c < z$, contradicting minimality of $z$.

Therefore, no positive integer solution exists.