<s>[INST] Find the four digit number? Find a four digit number which is an exact square such that the first two digits are the same and also its last two digits are also the same.
 [/INST] Given $\overline{aabb}=1100a+11b=k^2$, consider mod $4$:
$$k\equiv 0,1,2,3 \pmod{4} \\
k^2\equiv 0,1 \pmod{4}\\
1100a+11b\equiv 3b\equiv 0,1 \pmod{4} \Rightarrow b=0,3,4,7,8 \ \ \ \ \ \ (1)$$
Also, the last digit of $k^2$ can be:
$$b=0,1,4,5,6,9 \ \ \ \ \ \ (2)$$
Hence, from $(1)$ and $(2)$:
$$b=0 \ \ \text{or} \ \ 4.$$
And:
$$k^2=1100a+11b=11(100a+b)=11^2\cdot 9a+11(a+b) \Rightarrow a+b\equiv 0 \pmod{11}.$$
So, $\overline{aabb}=7744$.
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