$a+b+c=?$ by having $a^2+160=b^2+5$ and $a^2+320=c^2+5$

Question

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)

Answer 1

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 $b=18$.

Answer 2

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

Answer 3

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 $(b-a)$?