[Math] How many four digit numbers are perfect square whose first and last two digits are same

Question

I tried it by assuming the number as $\sqrt{1100a+11b}$ and than tried to find figure out perfect square but I am unable to approach further.

Answer 1

As

$$\sqrt{1100a+11b}=\sqrt{11(100a+b)}$$

To make the expression an integer, we may write

$$100a+b=11k^2,k\in{\Bbb{N}}$$

Also as $a,b\le9$, we have

$$100a+b=11k^2\le909$$

Furthermore, as $1000<1089=33^2$, we have $a\ge3$.

So $300\le100a+b=11k^2\le909$

And with some easy observation, we know that $6\le k<10$.

Obviously, only when $k=8,11k^2=704\implies a=7,b=4$.

So the number is $7744.$