Prove that there is not a single natural number $N$ with sum of digits equal to 15 that is the square of an integer.
Hint: If the sum of the digits is $15$, then $N\equiv 15\equiv 6\mod 9$. Can you show that $6$ is not a square $\bmod 9$?
Hint: If the sum of the digits is $15$, then $N$ is divisible by $3$, if $N$ is a square then it is also divisble by $9$, if $N$ is divisble by $9$, then the sum of digits...
Observe that the recursive sum of digit(R.D.) of any number of the form (9.a+b) where 0≤b<9 are same.
By recursive sum of digit, I mean continue taking the sum of digits until it becomes <10.
For example, $193^2 = 37249$ =>R.D. of $193^2$=R.D. of 37249 = R.D. of 25 = 7.
As 193≡4(mod 9) => $193^2≡4^2(mod\ 9)$ =>R.D. of $193^2$ is 7
Now R.D. of $(9a)^2$=9,
R.D. of $(9a±1)^2$=1,
R.D. of $(9a±2)^2$=4,
R.D. of $(9a±3)^2$=9 and
R.D. of $(9a±4)^2$=7.
But (R.D.) of 15 is 6.