<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:
So,  $a+b> a-b$ and $a+b=\dfrac{31\cdot5}{a-b}$
Either $a+b=155, a-b=?$  or $a+b=31,a-b=?$
Check which values of $a$ keep $c$ integer
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