Let $a\in \mathbb{Z}$ be such that $a^2=b^2+c^2$, where $b, c\in \mathbb{Z}^\setminus\{0\}$. Then $a$ cannot be written as
$pd^2$, where $d\in \mathbb{Z}$ and $p$ is a prime with $p\equiv 1\ \text{mod}\ 4$.
$pd^2$, where $d\in \mathbb{Z}$ and $p$ is a prime with $p\equiv 3\ \text{mod}\ 4$.
How to show that option 1 is incorrect but second is correct.
I know that modulo $4$, sum of any number is $0$ or $1$ modulo $4$. But don't get how to apply this result. Please help.
For 1, you can just show an example, like $5^2=3^2+4^2$
For 2, the fact that all squares are $0$ or $1 \bmod 4$ is what you want. If $d$ is odd, the left side is equivalent to $3 \bmod 4$, while the right is $0,1,$ or $2 \bmod 4$. If $d$ is even, the left side is $0 \bmod 4$, so both $b$ and $c$ are even. If there are any solutions to the equation, there is one with the smallest left side. If $d$ is even, we can divide all of $b,c,d$ by $2$ and have another solution. Keep going until $d$ is odd, then apply that case.