Sum of all $4$-digit squares of a specific form 
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.
 A: 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 $N=2809$.
A: EDIT: As the question has been edited to say $(ab)=3\times(cd)+1$ in stead of $(ab)\geq3\times(cd)+1$, this answer no longer answers the question.
The best I have is a very ugly brute force approach, I hope this is not the intended solution:
Let $N=(abcd)$ be a $4$-digit perfect square satisfying $(ab)\geq3\times(cd)+1$. Let $x\in\Bbb{N}$ be such that $x^2=N$, and note that we must have $x<100$.
It is clear that $(cd)$ is a quadratic residue modulo $100$ and it is at most $32$, so
$$(cd)\in\{0,1,4,6,16,21,24,25,29\}.$$
There are precisely $10$ solutions to $x^2\equiv0\pmod{100}$ with $0\leq x<100$, and it is obvious that for exactly $9$ of these the number $N=x^2$ satisfies the given inequality; these are
$$10,\ 20,\ 30,\ 40,\ 50,\ 60,\ 70,\ 80,\ 90,$$
and their squares sum up to $28500$.
For each quadratic residue $y\neq0$ there are precisely four solutions to $x^2=y$ yielding $32$ candidates for $x$. For each $y$ they are of the form
$$x_1=k\qquad x_2=50-k,\qquad x_3=50+k,\qquad x_4=100-k,$$
where $k\in\{1,2,3,4,5,11,18,23\}$. For solutions of the form $x_1$ we have $a=0$ in $x_1^2=N=(abcd)$, from which it easily follows that they cannot satisfy the inequality.
Solutions of the form $x_2$ work up to $k=2$, those of the form $x_3$ work up to $k=3$, and those of the form $x_4$ work up to $k=5$. These correspond to the perfect squares
$$2304,\ 2401,\ 2601,\ 2704,\ 2809,\ 9025,\ 9216,\ 9409,\ 9604,\ 9801,$$
which sum up to $59874$. This yields a sum of $28500+59874=88374$.
A: $\text{Given}:$
$N=x^2$
$N=\overline{abcd}$
$\overline{ab}=3\times\overline{cd}+1$

$\overline{ab}\lt100$
$3\times\overline{cd}+1\lt100$
$\overline{cd}\lt33$

$9\lt\overline {ab}$
$3\lt\overline{cd}\lt33$

$N=x^2$
$N\equiv0,1,4,5,6,9\pmod{10}$

$d=0,1,4,5,6,9$
$c=0,1,2,3$

$\overline{cd}=04,05,06,09,10,11,14,15,16,19,20,21,24,25,26,29,30,31$

Now, reducing possibilities. 
$\overline{cd}\neq10,20,30\quad[(10x)^2=100x^2]$
$\overline{cd}\neq5,15\;\;\;\;\quad\quad[(5x)^2=25x^2]$

So possible values of $\overline{cd}$ and corresponding values of $\overline{abcd}$:
$04,\quad1304$
$06,\quad1906$
$09,\quad2809$
$11,\quad3411$
$14,\quad4314$
$16,\quad4916$
$19,\quad5819$
$21,\quad6421$
$24,\quad7324$
$25,\quad7625$
$26,\quad7926$
$29,\quad8829$
$31,\quad9431$
Out of these, only $2809=53^2$ is a perfect square.
Therefore, answer is $2809$.
[Extremely Inelegant and Clumsy, but covers all possible solutions]
