Mathematical Reflections 2012, Issue 2 - Problem J227

Problem:
For a positive integer $N$ let $r(N)$ be the number obtained by reversing the digits of $N$.
Find all $3$-digit numbers $N$ such that $r^2(N)-N^2$ is the cube of a positive integer.

Proposed by Titu Andreescu.

Solution:
Let $N=100a+10b+c$ and $r(N)=100c+10b+c$ with $0 \leq a,b,c \leq 9$, $ac \neq 0$. Then
$$r^2(N)-N^2=(100c+10b+a)^2-(100a+10b+c)^2=99(c-a)[101(c+a)+20b].$$
Since $r^2(N)-N^2 > 0$, it must be $c > a$. If $r^2(N)-N^2$ is a cube, as $c-a \leq 8$, then $101(c+a)+20b \equiv 0 \pmod{121}$, i.e. $b-(a+c) \equiv 0 \pmod{121}$. Since $-17 \leq b-(a+c) \leq 6$, then $b=a+c$. So,
$$r^2(N)-N^2=11^3\cdot3^2(c-a)(c+a).$$
If $c-a=3$ then $c+a=2a+3$ and $5 \leq 2a+3 \leq 19$ must be an odd cube, contradiction. Likewise, $c-a=6$ yields $c+a=2(a+3)$, so that $a+3=2k^3$ where $k$ is a positive integer, contradiction. So, $c+a=3,6,9$.

(i) If $c+a=3$, then $a=1, c=2$ and it's easy to see that $r^2(N)-N^2=11^3\cdot3^3 = 33^3$.
(ii) If $c+a=6$, then $c-a=2(3-a)$ and $r^2(N)-N^2=11^3\cdot3^3\cdot2^2(3-a)$, so $a=1$ and $c=5$.
(iii) If $c+a=9$, then $c-a=9-2a$ and $r^2(N)-N^2=11^3\cdot3^4 (9-2a)$, but $9-2a=9k^3$  has no positive integer solution.

In conclusion, all the $3$-digit numbers which satisfy the required conditions are $132$ and $165$.