Exercise 2.1.58 (If $a^2 + 2b^2 = 2p$ has a solution, then also $a^2 + 2b^2 = p$.)

Proof. Since $a^2=2(p-b^2)$ is even, $a$ is also even (indeed, the square of an odd integer is odd).

Then $4{\left(\frac{a}{2}\right)}^2+2b^2=2p$, so $b^2=p-2{\left(\frac{a}{2}\right)}^2$, where $a∕2$ is an integer, is the sum of and odd integer with an even integer. Therefore $b^2$ is odd, thus $b$ is odd (the square of an even integer is even).

Multiplying by $2$ the equality $a^2+2b^2=2p$ , we obtain

and dividing by $4$,

This shows that if the equation $x^2+2y^2=2p$ has a solution $(a,b)$, then the equation $x^2+2y^2=p$ has also a solution, given by $(b,a∕2)$. □