For what values of $x$ is the number $121$ a square in base $x$?

This is a very easy question, but I still think it's pretty fun. Perhaps it's something to pass on to younger people who are taking algebra.


All of them (except base $2$ which can't have $121$.)

Added pedantry:

If you're ok with negative bases, the answer is all positive and negative bases except $-2$, $-1$, and $2$. Base $0$ and $1$ make no sense at all to me (and $-1$ very little) so I would also claim it doesn't hold for these, mostly because "holding" isn't really provably true or false, but also because 121 is not a valid number in those bases, as also it is not valid in base 2 and -2.

Also, I am considering integer bases only.

Why is it a square for these bases?

For whatever base $n$ you're working in, simply expanding the number $121$, it means $n^2 + 2n + 1$. And that is $(n+1)^2$ as a matter of algebra. And that is a square.

  • $\begingroup$ Base-1 excluded as well! $\endgroup$ – leoll2 Apr 22 '15 at 14:52
  • $\begingroup$ What about negative values of x? $\endgroup$ – blakeoft Apr 22 '15 at 14:54
  • $\begingroup$ 121 base minus 10 (ignoring any scruples about what a negative base even means) is 100 + (-20) + 1 which is 81, which is 9^2. So it doesn't seem to be an issue for bases -3 and lower. $\endgroup$ – Kate Gregory Apr 22 '15 at 15:07
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    $\begingroup$ @blakeoft For negative values of x, let n = abs(x). The formula is now n^2 - 2n + 1 which reduces to (n-1)^2 so the value will still be a square if x < -2. Therefore, the complete answer is Integers <-2 and > 2 $\endgroup$ – Engineer Toast Apr 22 '15 at 15:15
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    $\begingroup$ @meneldal The strange thing to me is that 121 doesn't exist in base 1 presumably because of the "2", but simultaneously the instances of "1" are ok. To be more specific, $n$ is not usually a legal symbol when writing a number in base $n$. I guess base 1 is an exception. $\endgroup$ – blakeoft Apr 24 '15 at 12:34

The base-$x$ number $121$ is a square

whenever $x\geq 3$.

For $x\geq 3$, we can multiply in base $x$: \begin{array}{ccc}&1&1\\\times&1&1\\\hline&1&1\\1&1&\\\hline1&2&1\end{array}We never needed to "carry", so this computation looks the same for every $x$.


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