Square which is Difference between Square and Square of Reversal

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Theorem

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

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


Proof

\(\ds 33^2\) \(=\) \(\ds 1089\)
\(\ds 65^2 - 56^2\) \(=\) \(\ds 4225 - 3136\)
\(\ds \) \(=\) \(\ds 1089\)


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 $x \ne y$:

\(\ds \sqbrk {xy}^2 - \sqbrk {yx}^2\) \(=\) \(\ds \paren {10 x + y}^2 - \paren {10 y + x}^2\)
\(\ds \) \(=\) \(\ds 100 x^2 + 20 x y + y^2 - 100 y^2 - 20 y x - x^2\)
\(\ds \) \(=\) \(\ds 99 \paren {x^2 - y^2}\)
\(\ds \) \(=\) \(\ds 3^2 \times 11 \paren {x - y} \paren {x + y}\)
\(\ds \leadsto \ \ \) \(\ds \paren {x - y} \paren {x + y}\) \(=\) \(\ds 11 n^2\) for some integer $n$

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:

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

$\blacksquare$


Sources

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