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This question is inspired by Google's recent programming competition (modified slightly for ease of exposition).

For a given $n$, one of the problems was to find all positive "fair" integers $k$ less than $n$, where $k$ is "fair" if

  1. $k$ is a palindrome (in base 10, no leading zeros)
  2. $k^2$ is also a palindrome.

One first result is that if $k$ is a palindrome, then $k^2$ will involve no carrying if and only if the sum of $k$'s squared digits is less than ten. Therefore all palindromes $k$ with sum of squared digits less than ten are "fair."

But can there be sporadic "fair" numbers? Palindromes $k$ where computing $k^2$ involves some carrying, but by pure chance $k^2$ still ends up being a palindrome?

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up vote 1 down vote accepted

During the competition, I found the first few 'fair and square' numbers, then searched for them in Sloane's encyclopedia of integer sequences. Its page suggests the 'sum of squares of digits is less than 10' condition is necessary as well as sufficient, but doesn't give a proof. It's not obvious to me either.

Edit: Google posted a proof by contradiction at

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I've tried solving the same problem yesterday. I managed to brute-force my way to finding all fair and square (palindromes whose square root is a palindrome) numbers from 1 to $10^{14}$:

1, 4, 9, 121, 484, 10201, 12321, 14641, 40804, 44944, 1002001, 1234321, 4008004, 100020001, 102030201, 104060401, 121242121, 123454321, 125686521, 400080004, 404090404, 10000200001, 10221412201, 12102420121, 12345654321, 40000800004, 1000002000001, 1002003002001, 1004006004001, 1020304030201, 1022325232201, 1024348434201, 1210024200121, 1212225222121, 1214428244121, 1232346432321, 1234567654321, 4000008000004, 4004009004004

My solution was correct, and the second dataset was solved. But I couldn't find a proper way to calculate all fair and square numbers up to $10^{100}$.

I showed this to my wife this morning, and she noticed an interesting pattern of numbers within my list:

121, 10201, 1002001, 102030201, 10000200001, 1000002000001
484, 40804, 4008004, 400080004, 40000800004, 4000008000004
12321, 1002003002001,

Some fair and square numbers re-appear with space padding. Let's try beyond $10^{14}$. Adding some zeros to $1020302030406040302030201$, whose square root is $1010100010101$ - a palindrome!

Wish I had my wife with me when I solved this yesterday.

I don't have a mathematical explanation for this phenomena, but I guess that for some reason, every fair and square number beyond a certain boundary can be built by adding zeros to a smaller palindrome.

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The first 412 of these are tabulated at

All the ones listed have no digit exceeding 2, and generally have plenty of zeros, so I'm guessing there aren't any sporadics in the list. Of course, that's not a proof....

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+1 Nice approach - find the root instead of the number itself. – Adam Matan Apr 14 '13 at 6:20

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