Find all pairs of positive integers (a,b) such that a and b each have two digits and that 100a+b and 201a+b are perfect squares with 4 digits each.
Here's one "brute strength" way to do it- all perfect squares with four decimal places are between $\displaystyle 31^2$ and $\displaystyle 100^2$. There are 68 such numbers. square every number form 32 to 99. "a" will be the first two digits and "b" will be the last two. Then check to see if 201a+ b is also a perfect square. For example, if n= 50, then $\displaystyle n^2= 2500$. a= 25 and b= 0. 201a+ b= 5025. $\displaystyle \sqrt{5025}= 70.88$ which is not an integer so 201a+ b is not a perfect square.