Perfect squares always have recursive sum of their digits(e.g. $361\rightarrow 10 \rightarrow 1$) as either $1, 4, 7$ or $9$. But if the perfect square is square of a prime number, except of $3$ . This recursive sum is always $1,4$ or $7$. Its never $9$ for squares of prime numbers after the square of prime number $3$. Any mathematical reason?
2026-03-27 23:23:37.1774653817
Recursive sum of squares of prime numbers
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If the recursive sum of the digits of any integer is $9$, the number is always multiple of $9$. If the number is a perfect square, say $n^2$, then $n$ must be a multiple of $3$. So if you square a prime distinct from $3$, the recursive sum of the digits can not be $9$.
The fact of that the recursive sum of the digits of every positive multiple of $9$ is $9$ can be easily proven with modular arithmetic:
Take any natural number $n$ with $r+1$ digits. Its decimal representation is: $$n=\sum_{k=0}^r a_k10^k$$ But, since $10\equiv1\pmod9$, $$n\equiv\sum_{k=0}^r a_k10^k\equiv\sum_{k=0}^r a_k \pmod 9$$
If $n$ is multiple of $9$, then so is the sum of its digits. And then, so is the sum of the sum of its digits, etc.
Every time you add the digits of a number with more than $1$ digit, you get a strictly lesser number, so you must eventually get $9$, after a finite number of sums.