2026/030) How can we solve this equation $a^-ab+b^2=1 such that a and be are positive integers?

Multiply by 4 on both sides to get

$4a^2 -4ab + 4b^2 = 4$

Or $4a^2 - 4ab + b^2 + 3b^2 = 4$

Or $(2a -b)^2 + 3b^2 = 4$ 

Looking at integer solutions $2a - b = 1, b = 1$ as b above 1 becomes larger

Giving $a = 1, b = 1$