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The other day I thought of this question:
Is it possible to find a set of palindromic numbers such that the square of them is itself palindromic? In other words:

$A = \{a : a^{2}=b$ where a,b are palindromic integers}

I'm not a mathematician so I don't really know where to begin!

Edit: Thanks to all who responded so quickly! My question now is whether is possible to find ALL possible integers that belongs to such set. Would it be possible to find a rule for that set?

Davide M
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    A subset of $A$ is $B={10^n+1 : n\in\Bbb{N}}$ – kingW3 Jun 21 '17 at 17:08
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    OEIS sequence A057135 – Robert Israel Jun 21 '17 at 17:14
  • As mentioned in OEIS sequence A057135, this consists of all palindromes such that the sum of squares of the digits is less than 10. This lets you quickly generate all such numbers with a given number of digits. Except for $3$, the number must consist only of digits $0$, $1$ and $2$. A single $2$ would occur as the central digit, with at most two 1's on each side (e.g. $10100200101$). Or there could be two $2$'s, but then at most one $1$, so the pattern here would be $2 0\ldots0 2$ or $2 0\ldots 010\ldots 02$. Otherwise, it's all $0$'s and $1$'s, with at most $9$ $1's$. – Robert Israel Jun 21 '17 at 18:14

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If you keep to the digits $0, 1,$ and $2$, and include plenty of $0$'s you can make lot of examples.

$$1002001^2 = 1004006004001$$

$$212^2 = 44944$$

Shaun
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B. Goddard
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