Find all $a,b,c \in \Bbb Z$, such that $a^2 +31b^7=c^2$

Question

Find all $a,b,c\in\mathbb{Z},$ such that $a^2 +31b^7=c^2.$ Here is my approach: Factoring $c^2-a^2=31b^7$ we get, $(c+a)(c-a)=31b^7$
$c-a=b^7$
$b=1$

$c+a=31$
$2c=32$
$c=16$
$a=15$

$c-a=31$
$c+a=b^7$

$2c= b^7+31$
$c=(b^7+31)/2$
$a=b^7-31/2$

$b^7=2n-1$
$b=2n-1$

$(15,1,16)((b^7-31)/2,b,(b^7+31)/2) (b=2n-1)$

Is my solution true?

Answer 1

If $a,b,c\in\Bbb{Z}$ are such that $a^2+31b^7=c^2$ then also $$31b^7=c^2-a^2=(c+a)(c-a).$$ This yields a factorization $31b^7=uv$, with $u=c+a$ and $v=c-a$, such that $u\equiv v\pmod{2}$.

Conversely, let $b$ be any integer and $31b^7=uv$ a factorization such that $u\equiv v\pmod{2}$. Then $$a:=\frac{u-v}{2}\qquad\text{ and }\qquad c:=\frac{u+v}{2},$$ are integers satisfying $a^2+31b^7=c^2$.