find a number $aabb$ such that is full square

Question

I know that I can write $aabb=1000a+100a+10b+b$, $a,b\in \{0,1,2,3,4,5,6,7,8,9\}, a\neq 0$, so $aabb=1100a+11b=m^2$, where $m\in \mathbb N$, I notice that $m^2=11(100a+b)$ so I need to find some number that $11\mid m^2$ and $11\mid m$,$1100<m^2<9999$ or $33<m<99$ so I just put number in calculator and 88 is answer, so $aabb=7744$, but is there any other better solution?

Answer 1

We have that

$$11\mid (100a+b) \implies (a+b)\equiv 0 \pmod{11},$$

and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.