2022/071) Find integers a,b,c such that $a^2-b^2 = 24$ and $b^2-c^2= 16$

We ahve $a^2-b^2 = 24\cdots(1)$

$b^2-c^2 = 16\cdots(2)$ 

Because $a^2-b^2 = 24$ so both a and b are odd or both ae even,

So we have a = b + 2n for some n

To put an upper bound on be a = b+ 2 beause a is at least 2 more than b 

So we have $a^2-b^2 = (b+2)^2 - b^2  = 24$

Or $2b + 4 = 24$ or $b= 5$

For the lower bound $b^2 =17$ so $b ge 5$

So we get b = 5 and putting the values 

So we get $a= 7, b = 5, c= 3$