Square which is Difference between Square and Square of Reversal

Theorem
$33^2 = 65^2 - 56^2$

This is the only square of a $2$-digit number which has this property.

Proof
Let $\sqbrk {xy}$ be a $2$-digit number such that $x \le y$ and $\sqbrk {xy}^2 - \sqbrk {yx}^2$ is a square of a $2$-digit number.

The case $x = y$ gives the solution $\sqbrk {xy}^2 - \sqbrk {yx}^2 = 0$, which is not a square of a $2$-digit number.

For $x \ne y$:

By Euclid's Lemma for Prime Divisors, one of $x - y, x + y$ must be divisible by $11$.

Hence from Absolute Value of Integer is not less than Divisors, either $x - y$ or $x + y$ must be greater than or equal to $11$.

Since $x - y < x < 9$, $x - y$ cannot be a multiple of $11$.

From $1 = 1 + 0 \le x + y < 9 + 9 = 18$, we have that $x + y = 11$.

This implies that $x - y$ is a square number.

$x + y = 11$ gives $\tuple {x,y} = \tuple {6,5}, \tuple {7,4}, \tuple {8,3}, \tuple {9,2}$ as the possible solutions.

Among these solutions, only $\tuple {6,5}$ has a difference of a square number: $1$.

Therefore the only square of a $2$-digit number expressible as a difference between a square and the square of its reversal is:


 * $65^2 - 56^2 = 33^2$.