Fairer and Squarer

Let the two squares be $10000 \leq M^2 < N^2 \leq 99999$ . We conclude immediately that $100 \leq M,N \leq 316;\ \ \ \ \ \ \begin{cases} 200 \leq N+M \leq 632 \\ N-M \leq 216\end{cases}$ Now let $a$ and $e$ be the first and last digit of $N$ ; then $N^2 - M^2 = 10000(a-e) + (e-a) = 9999(a - e) =: 9999h.$ Using the factorization $9999 = 3^2\cdot 11\cdot 101$ , we conclude that $(N+M)(N-M) = 3^2\cdot 11\cdot 101\cdot h$ with $h = 1, \dots, 8$ . Answers to the problem must be found among solutions of this equation. Note also that $N+M$ and $N-M$ must have the same parity (odd or even). If they are both even, we necessarily have $h = 4$ or $8$ .

The prime factor $101$ must be either a divisor of $N+M$ or of $N-M$ .

Case 1. $101$ is a divisor of $N+M$ . This implies $N+M = 202, 303, 404, 505,\ \text{or}\ 606$ .

If $N+M = 303$ or $606$ , then we see that $3\cdot 11 = 33$ must be a divisor of $N-M$ . Knowing that $N-M \leq 216$ and that $N+M$ and $N-M$ have the same parity, this leaves the following possibilities:

$\begin{array}{cc|cc|cc|l} N+M & N-M & N & M & N^2 & M^2 & \\ \hline 303 & 33 & 168 & 135 & 28224 & 18225 & \text{no match} \\ 303 & 99 & 201 & 102 & 40401 & 10404 & \text{SOLUTION (already known)} \\ 303 & 165 & 234 & 69 & \dots & \dots & M\ \text{too small} \\ \hline 606 & 66 & 336 & \dots & \dots & \dots & N\ \text{too large} \\ 606 & 132 & & & & & \text{idem} \\ 606 & 198 & & & & & \text{idem} \\ \hline \end{array}$

If $N + M$ is another multiple of $101$ , we conclude that $3^2\cdot 11 = 99$ must be a divisor of $N-M$ . This means that $N-M = 99$ if $N+M$ is odd, and $198$ if $N+M$ is even.

$\begin{array}{cc|cc|cc|l} N+M & N-M & N & M & N^2 & M^2 & \\ \hline 202 & 198 & \dots & 2 & \dots & \dots & M\ \text{too small} \\ 404 & 198 & 301 & 103 & 90601 & 10609 & \text{SOLUTION (already known)} \\ 505 & 99 & 302 & 203 & 91204 & 41209 & \text{SOLUTION (already known)} \\ \hline \end{array}$

Case 2. $101$ is a divisor of $N-M$ . This implies $N-M = 101$ or $202$ . It follows that $3^2\cdot 11 = 99$ is a divisor of $N+M$ . This gives the following possibilities:

$\begin{array}{cc|cc|cc|l} N+M & N-M & N & M & N^2 & M^2 & \\ \hline 297 & 101 & 199 & 98 & \dots & \dots & M\ \text{too small} \\ 396 & 202 & 299 & 97 & \dots & \dots & M\ \text{too small} \\ 495 & 101 & 298 & 197 & 88804 & 38809 & \text{no match} \\ 594 & 202 & 398 & \dots & \dots & \dots & N\ \text{too large} \\ \hline \end{array}$

Conclusion. Beyond the three solution already shown in the problem, we found $\boxed{\text{no other pairs}}$ of five-digit perfect squares with this property.

Mathematica

For[x = 100, x < 317, x++, For[y = x + 1, y < 317, y++, If[IntegerDigits[x^2][[{1, 5}]] == IntegerDigits[y^2][[{5, 1}]] && IntegerDigits[x^2][[{2, 3, 4}]] == IntegerDigits[y^2][[{2, 3, 4}]], Print[x, " ", y]]]]

102 201
103 301
203 302