The conjecture:

The following relation, $$r(x)=d(x)(\sqrt{x}+1)$$ where $r(x)$ is the reverse of $x$ and $d(x)$ is the sum of the digits of $x$ (not the digital root, just the sum) and is a perfect square and $x \in \mathbb{N}$ is satisfied only by 36. Prove or disprove this.

We have tried to solve it, but have made no fruitful progress whatsoever. The question itself was posed purely as a recreational challenge.


The number of digits of $x$ is the base $10$ log of $x$, rounded up. You can bound $d(x)$ by $9 \lceil\log_{10}(x)\rceil$ which means the right side grows much more slowly than the left. Find a value for $x$ such that any greater number has the left greater than the right, then just try all the $x$ below that. The left is an integer, so you only need to try $x$s which are perfect squares. The better an upper bound you put on $x$ the less work you have to do in the try stage.

As pointed out in a comment, this argument fails if $x$ has trailing zeros because $r(x)$ can be much less than $x$. If $x$ has trailing zeros, there must be an even number or $x$ will not be square, so we can write $x=y10^{2k}$ where $y$ does not end in zero. The equation becomes $$r(y)=d(y)(10^k\sqrt y+1)$$ So far I don't see how to put a bound on $y$ like the first paragraph does on $x$

  • $\begingroup$ But how can you say that the left grows faster than the right? $\endgroup$ – Mainak Roy Mar 23 '18 at 15:09
  • $\begingroup$ Because $r(x)$ can't be much smaller than $x$-only a factor $10$. The left grows as $\sqrt x \log_{10}(x)$. Say $x$ is six digits. The left side is at least $100000$. The right is at most $54\sqrt {999999}\lt 54000$. The inequality only gets larger as $x$ grows. You can be more careful and reduce the bound, but we already only have $1000\ x$s to try. $\endgroup$ – Ross Millikan Mar 23 '18 at 15:13
  • $\begingroup$ But what about numbers like say $441$? Then $r(x)=144$. Or $100$, so $r(x)=1$? $\endgroup$ – Mainak Roy Mar 23 '18 at 15:19
  • $\begingroup$ $441$ is less than the limit, so is no problem. We would have to check it. Numbers with trailing zeros are a problem because now $r(x)$ can be small. I'll have to think on that one. $\endgroup$ – Ross Millikan Mar 23 '18 at 15:22
  • $\begingroup$ I missed that $d(x)$ is supposed to be square in the problem statement. Does that change anything? $\endgroup$ – Mainak Roy Mar 30 '18 at 14:20

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