For every odd prime number $p$ there exists a unique natural number $x$ such that $p^2 + x^2$ is a square

Question

I have to prove that for every odd prime number $p$ there exists exactly one natural number $x$ such that $p^2 + x^2 = n^2$

I have no clue how to attempt this one and im thankful for hints.

Answer 1

Hint: $p^2=n^2-x^2=(n-x)(n+x)$ which leads to $n-x=1$.

Alternatively, $p$ - odd then $p^2$ is also odd. I.e. $p^2=2x+1 \Rightarrow p^2+x^2=x^2+2x+1=(x+1)^2$

Answer 2

The parametrization of Pythagorean triples $$(x,y,z)=(m^2-n^2,2mn,m^2+n^2)$$ gives a proof because $m^2-n^2=(m+n)(m-n)$ so making $m=n+1$ we have $p=2n+1$ and we get $$(2n+1)^2+(2n(n+1))^2=(n^2+(n+1)^2)^2$$ which is easy to verify. You finish making $$\begin{cases}p=2n+1\\x=2n^2+2n\\a=2n^2+2n+1\end{cases}$$