Find the sum of all $4$-digit squares $N=(\overline{abcd}) $ for which $(\overline{ab}) =3(\overline{cd}) +1$.
(The notation $n= (\overline{ab}) $ means that $n$ is a 2-digit number and its
value is given by $n= 10a + b$)
I came across this question in a Math Olympiad Competition and did not have the slightest clue on how to solve it. Can anyone help? Thanks.
Best Answer
Write $N=x^2$. Then $x^2=N=\overline{abcd}=100\overline{ab}+\overline{cd}=301\overline{cd}+100 \equiv 10^2 \pmod{7*43}$
Thus $x \equiv \pm 10 \pmod{7}$ and $x \equiv \pm 10 \pmod{43}$.
Thus $x \equiv 10, 53, 248, 291 \pmod{301}$. Note $32 \leq x \leq 99$. Thus $x=53$ and $N=2809$.