Cfrgtkky

What is the method for finding numbers $a$ and $b$?

Recall the divisibility criteria for 11: a number is divisible by 11 if and only if the alternating sum of its digits is divisible by 11.

Hint. Hence the number $overline{aabb}$ is divisible by $11$ and
$overline{aabb}=11cdot overline{a0b}$.

So if $overline{aabb}$ is a non-zero perfect square then also $overline{a0b}$ is divisible by 11, which means that $a=11-b$.

Can you take it from here?

P.S. For the final answer see Erick Wong's comment below.

Let $N=overline{aabb}=11(100a+b)$. For $N$ to be a square, we need $100a+bequiv a+bequiv0$ mod $11$. Since $0le a,ble9$, we must have $b=11-a$ (unless $a=b=0$), in which case $N=11(99a+11)=11^2(9a+1)$. This implies $N/11^2=9a+1equiv a+1$ mod $8$. Since the only squares mod $8$ are $0$, $1$, and $4$, we need only consider $a=3$, $7$, and $8$. Of $9a+1=28$, $64$, and $73$, only $a=7$ corresponds to a square. Thus $N=7744$ is the only possibility (aside from $0000$).

$$overline{aabb}=11(100a+b)$$

So, we need $11mid(100a+b)iff11mid(a+b)$

Now $0le a,ble9$

Again, $xequiv0,pm1,pm2,pm3,pm4,5pmod{10}$

$implies x^2equiv0,1,4,9,6,5$

Consider the possible final two digits modulo $100$ - show you can't have a square ending in $11, 55, 66, 99$ for example, which leaves very few options.

By divisibility of $11$, we have
begin{align*}
overline{aabb} &= 11k \
n^2 &= 11k \
k &= 11 c^2 \
overline{aabb} &= 11^2c^2 \
&= (overline{cc})^2
end{align*}

where $1le a,$ $b$, $cle 9$

begin{align*}
11c times 11c &= 1100a+11b \
11 times 11 c^2 &= 11(100a+b) \
11 c^2 &= 100a+b \
&= 99a+a+b \
a+b &= 11(c^2-9a) \
end{align*}

Note that $2le a+b le 18$,
$$
left{
begin{array}{rcl}
a+b &=& 11 \
c^2-9a &=& 1
end{array}
right.$$

Now
begin{align*}
9a &= c^2-1 \
&= (c+1)(c-1)
end{align*}

By $c-1<9$ and $3a-3 ne 2$,
$$
left {
begin{array}{rcl}
9 &=& c+1 \
a &=& c-1
end{array}
right.$$

Therefore
begin{align*}
c &=8 \
a &=c-1 \
&=7 \
b &=11-a \
&=4
end{align*}

That is $$7744=88^2$$