$a+b+c=?$ 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)

We know that: $$ (b-a)(a+b)=31\times 5=155$$ as both $31$ and $5$ are prime numbers we can say either $b-a=5, a+b=31$ OR $b-a=1, a+b=155$. (becasue $a+b>b-a$)


*

*$b-a=5, a+b=31$, thus: $$a=13 , b=18$$
On the other hand, we have:
$$a^2+320=c^2+5$$
$$\to c^2=13^2+320-5 \to c=22$$

*$b-a=1, a+b=155$, thus: $$a=77 , b=78$$
On the other hand, we have:
$$a^2+320=c^2+5$$
$$\to c^2=77^2+320-5 \to c\approx 79.02 \not\in \Bbb Z$$

Hence the only correct answer is $a=13$ and $b=18$.

