Question

help
There are positive integers a, b, and c that satisfy this system of two equations:
{c(suared)-a(squared)-b(squared)=101
{ab=72
What is the value of a + b + c?

Answer 1

$$c^2-a^2-b^2=101\\c^2-a^2-b^2-2ab=101-144\\c^2-(a+b)^2=-43\\(a+b)^2-c^2=43\\(a+b+c)(a+b-c)=43$$note however \(43\) is prime i.e. \(43=43\times1\) hence \(a+b+c=43\)

Answer 2

if ab=72 and a and b are both integers, they have to be factors of 72 (factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 and a lot of these will be too big to work in the first equation).

I used trial and error, starting from the middle factors 8 and 9, and used
\[\Large c^2 - a^2 - b^2 = 101\]so \[\Large c^2 = 101+a^2 + b^2\]and just looked for a perfect square by plugging in a and b.

8 and 9 failed, 12 and 6 failed, 4 and 18 works as it gives c^2 = 441, so c=21.

a+b+c = 4+18+21 = 43.