For a Diophantine equation $x^2+py^2=z^2$ show that $z$ is necessarily odd.

Question

For a Diophantine equation $x^2+py^2=z^2$ where $p$ is a prime of the form $p\equiv 1(mod4)$ and $(x,y,z)=1$.

Show that $z$ is necessarily odd.

Answer 1

Suppose to the contrary that $z$ is even.

If $y$ is even, then $x^2$ is even, so $x$ is even, contradicting the fact that $\gcd(x,y,z)=1$. Similarly, if $x$ is even, then $py^2$ is even, and since $p$ is odd $y^2$ is even, and therefore $y$ is even, and again $\gcd(x,y,z)\gt 1$.

So $x$ and $y$ are both odd. Thus $x^2\equiv 1\pmod{4}$ and $y^2\equiv 1\pmod{4}$, and therefore $x^2+py^2\equiv 2\pmod{4}$. This is impossible, since any even square is congruent to $0$ modulo $4$.

Answer 2

For the equation:

$$x^2+py^2=z^2$$

There is a solution.

$$x=t^2-pk^2$$

$$y=2tk$$

$$z=t^2+pk^2$$

Means it is necessary to rewrite the equation in the form. $p=4s+1$

$$x^2+(4s+1)y^2=z^2$$

This is possible if the number of $x,y - $ odd.

Means: $$z=\frac{t^2+(4s+1)k^2}{2}$$

and the numbers $t,k - $ always odd. Imagine the numbers.

$$t=2t+1$$

$$k=2k+1$$

We substitute in the formula and find out what the parity of the number $z$.

$$z=2s(2k+1)^2+2(t^2+k^2)+2(t+k)+1$$

This number is always odd! With the high value of the coefficient.