4. $[\mathbf{2 0}]$ Find all 4-digit integers of the form $a a b b$ (when written in base 10) that are perfect squares.
Answer: 7744 Let $x$ be an integer such that $x^{2}$ is of the desired form. Then $1100 a+11 b=x^{2}$. Then $x^{2}$ is divisible by 11 , which means $x$ is divisible by 11 . Then for some integer, $y, x=11 y$. Then $1100 a+11 b=11^{2} y^{2} \Rightarrow 100 a+b=11 y^{2}$. This means that $100 a+b \equiv 0(\bmod 11) \Rightarrow a+b \equiv 0(\bmod 11)$. Because $a$ and $b$ must be nonzero digits, we have $2 \leq a, b \leq 9$, so we can write $b=11-a$.
Replacing $b$ in the equation derived above, we obtain $99 a+11=11 y^{2} \Rightarrow 9 a+1=y^{2}$. We check the possible values of $a$ from 2 to 9 , and only $a=7$ yields a perfect square. When $a=7, b=4$, so the only perfect square of for $a a b b$ is 7744 .