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$12^2 = 144$

Now reverse the digits and square it: $21^2 = 441$

The digits in the result have also gotten reversed.

The same is true for $13$: $13^2 = 169$, $31^2 = 961$

As far as I know, this does not hold true for any other number (except, trivially, $0$ through $11$).

Why? Can it be proven? The only thing that occurs to me is that, starting with the next number, $14$, the square of its reverse, $41$, is greater than $1000$. But I don't know why this would make any difference.

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    Also worth noting that the sequence of such is [A140212](https://oeis.org/A140212) – 2017-02-06

1 Answers 1

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In base $B$, $(mB+n)^2 =m^2B^2+2mnB+n^2 $ and $(nB+m)^2 =n^2B^2+2nmB+m^2 $.

If all of $m^2, n^2,$ and $2mn$ are less than $B$, then there will be no carrying are the results will be as you have described. If any of them are at least $B$, then there will be carrying and the symmetry will be gone.

For $B=10$, the possibilities with $m \le n$ are $(m, n) =(1, 1), (1, 2), (1, 3) ,(2, 2) $.

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    Terrific. It never occurred to me to think it out like this. – 2017-02-06