Fair and Square

Let the two squares be $N^2 > M^2$ . Since they have four digits, it is clear that $99 \geq N > M \geq 32$ .

If $a > d > 0$ are the first and last digits of $N^2$ , then we have $N^2 - M^2 = (1000a - d) - (1000d - a) = 999(a - d).$ Define $h = a - d$ ; we have $(N + M)(N - M) = 999h,\ \ \ 1 \leq h \leq 8.$ We know that $\begin{cases} N - M \leq 99 - 32 = 67 \\ 64 \leq N + M \leq 198 \end{cases}$ Moreover, the prime factor decomposition $999 = 3^3\cdot 37$ . This means that either $N + M$ or $N - M$ is a multiple of 37. This greatly reduces the number of situations we must consider. The above equations are necessary conditions for finding the solution, but not sufficient . We must therefore carefully check each solution to see if it really works.

Case 1 . If $N + M$ is a multiple of 37, then it is at least $2\times 37 = 74$ and at most $5\times 37 = 185$ .

In the special situation $N + M = 111 = 3\times 37$ , $N - M = 3^2 h$ is a multiple of 9. Since $N + M$ and $N - M$ are of equal parity, $h$ must be odd; we have the possibilities

• $h = 1$ , $N - M = 9$ , $N = (111+9)/2 = 60$ , $M = (111-9)/2 = 51$ ; this is not a solution because $N^2$ may not end in zero.

• $h = 3$ , $N - M = 27$ , $N = (111+27)/2 = 69$ , $M = (111-27)/2 = 42$ . This is the solution we look for: $69^2 = \boxed{4761},\ \ \ 42^2 = 1764.$

• $h = 5$ , $N - M = 45$ , $N = 78$ , $M = 33$ . However, $78^2 = 6084$ and $33^2 = 1089$ ; this is not a solution.

Otherwise, $N - M$ is a multiple of $3^3 = 27$ , i.e. equal to either $27$ or $54$ . Again, $N + M$ and $N - M$ must have equal parity. Thus

• $N + M = 2\cdot 37 = 74$ , $N - M = 54$ , $N = 64$ , $M = 10$ . This would make $M^2$ end in zero, which is not possible.

• $N + M = 4\cdot 37 = 148$ , $N - M = 54$ , $N = 101$ is too large.

• $N + M = 5\cdot 37 = 185$ , $N - M = 27$ , $N = 106$ is too large.

Case 2 . If $N - M$ is a multiple of 37, then it must be equal to 37, and $N + M = 27h$ . Equal parity requires that $h$ is odd. Thus

• $N - M = 37$ , $N + M = 3\times 27 = 81$ , $N = 59$ , $M = 22$ . But this would give $M^2$ only three digits.

• $N - M = 37$ , $N + M = 5\times 27 = 135$ , $N = 86$ , $M = 49$ . However, then $M^2$ would end in 1 and $N^2$ is clearly greater than 2000. This is not a solution.

• $N - M = 37$ , $N + M = 7\times 27 = 189$ , $N = 113$ is too large.

Conclusion : We found only one solution, the pair $69^2 = \boxed{4761},\ \ \ 42^2 = 1764.$