Integers that squared have the same last two digits

Question

I need to find the integers that when squared they maintain the las two digits, I've started like this: being b and a the last two digits of the number $(10b+a)²\equiv 10b+a \mod(100) \Rightarrow 20ab + a² \equiv 10b+a \mod(100) $ I don't know how to continue this, any idea?

Answer 1

HINT: Start by checking which digits $a$ can be; there are only four possibilities. Then substitute each of those possibilities into your congruence

$$20ab+a^2\equiv 10b+a\pmod{100}$$

and solve for $b$; note that there may be no solution in some cases.