We have \(\overline{aabb}=11\times\overline{a0b}=11\times\left(100a+b\right)\)   

11 divide by 4 give we remainder 3. A square number divide by 4 give me remainder 0 or 1 

==> Look in 100a + b . \(100a⋮4\) so if b : 4 give we re. 1 or 2 the 11(100a + b ) gives re. 3 or 2 (no!)

So b : 4 must gives re. 3 or 4 , b is a digit ==> b = { 3 ; 4 ; 7 ; 8}. Square never have last dg. is 3;7 and 8 , so b =4

With sentences above we get 100a + b (100a + 4) divide by 3 gives re. 2 or 0. Remember b = 4 = 3 + 1, so a : 3 gives re. 1or 2. Grouping a we get a = {1 ; 2 ; 4; 5; 7; 8}. Try each a you'll find a=7

That number is 7744. A sq. of 88

Do you think its