Given: a/b=x^2, a/b+5=y^2, a/b-5=z^2

Write (a+5b)/b=y^2, but y^2 is rational, say (y1^2/y2^2). Multiply: (a+5b)=b(y1^2/y2^2). 

But (a+5b) is an integer, so b must cancel (y2^2); so y2^2=b, meaning b is a square; and if b is square, so too must a be. 

This gives us the new equations, with slight modification of variables: 

a^2/b^2=x1^2/x2^2, or simply a^2=x^2 (clearly, b^2 must be less than a^2, or z^2 would be negative) 

(a^2+5b^2)=y^2, 

(a^2-5b^2)=z^2; 

These are 3 squares in arithmetic progression, so congruum numbers will apply. The sequence (Sloane A256418) starts 24, 96, 120, 216, 240, 336, 384, 480, 600, 720, and the smallest of these divisible by 5 leaving a square is 720, since 720/5=144. Thus, b^2 is 12^2.

This gives the updated equations: (a^2+720)=y^2, (a^2-720)=z^2, where a=41, y=49, and z=31.

Plugging back these results to the original problem: (a/b)=(41^2/12^2), a/b+5=49^2/12^2, and a/b-5=31^2/12^2.