The end feels a bit unnecessary. Once you know that a² + (b+2)² = 100, you simply observe that the only way to write 100 as a sum of positive squares is 8²+6² (this is quickly seen by exhaustion). Thus, (8, 6-2) and (6, 8-2) are the only solutions.

Let c=b+2, then a^2+c^2=100. Since it's symmetric, you only need to verify 5 cases (1,2,3,4,5) instead of 7 cases. You can also save the efforts of proving b

All values of a between (-10,10) will give two solution for b between [-12,8] and a=±10 will give b=-2 as solution.

a^2 + (b+2)^2 = 100 if a >= b+2 then 7 < sqrt(50)

As I solved the equation from the thumbnail, I didn't see the restriction to positive integers, so got more solutions, 12 in fact. a²+b²+4b+4=100 a²+(b+2)²=100=4×5² a and b are both even, as the square of an odd number is of the form 4n+1 (a/2)²+(b/2+1)²=5² As the only solutions in positive integers to x²+y²=5² are x=3, y=4 & vice versa, Solutions are |a/2|=0, |b/2+1|=5: a=0, b=8, -12 |a/2|=3, |b/2+1|=4: a=±6, b=6, -10 |a/2|=4, |b/2+1|=3: a=±8, b=4, -8 |a/2|=5, |b/2+1|=0: a=±10, b=-2