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 $x$