Sum of digits of a perfect square in other bases
I noticed that in base ten, the sum of digits of $n$ is equivalent to $n\pmod{9}$, and if a number in decimal is perfect square, then the sum of digits of that number must be a quadratic residue $\pmod{9}$, such like this one.
I think that the decimal is the best radix to find perfect squares as if the sum of digits of a number is divisible by $3$, then the sum of digits of that number must be divisible by $9$, as if $3\mid n$, then $3^{2}\mid n$. This comes via divisibility rule.
I don’t know if this is true for other bases that is form of $a^{b}+1$.
For example, if the sum of digits of a number in base twenty-six is either $15$ or $22$, then is that number not a perfect square, similar when the sum of digits of a number in decimal is $6$ or $8$?
Also, if the sum of digits of a number in base twenty-eight is $18$, then that number isn’t a perfect square also?
The divisibility rule about summing the digits to get the remainder on division by $9$ applies in all other bases $b$ to get the remainder on division by $b-1$. If you express a number in base $26$ and sum the digits you will get the remainder on division by $25$. The sum must be a square $\pmod {25}$ for the number to possibly be a square. As neither $15$ nor $22$ is a square $\pmod {25}$ no number with a digit sum of $15$ or $22$ base $26$ can be either.