<s>[INST] $a+b+c=?$ by having $a^2+160=b^2+5$ and $a^2+320=c^2+5$ My question:
$a+b+c=?$ by having $a^2+160=b^2+5$ and $a^2+320=c^2+5$.
My work so far:
$a^2+160=b^2+5\Rightarrow (b-a)(a+b)=155=31\times 5$
$a^2+320=c^2+5\Rightarrow (c-a)(c+a)=315=5\times3^2\times 7$
And now, I'm stuck.
($a,b,c$ are a members of $\mathbb Z$ and are positive)
 [/INST] Hint:
$$(b+a)(b-a)=31\times 5$$
Since $a,b$ are integers, and $5,31$ are prime numbers, what is the value of $(b+a)$ and $(b-a)$?
</s>