IRL_ABooklet_2023

The squares modulo $9$ are $\{0,1,4,7\}$. Any integer is congruent modulo $9$ to the sum of its digits (as powers of $10$ are congruent to $1$ modulo $9$). Therefore any totally square integer is congruent modulo $9$ to one of $\{0,1,4,7\}$. Taking all combinations of pairs, the sum of two totally square integers is congruent modulo $9$ to one of $\{0,1,2,4,5,7,8\}$. Positive integers congruent to $3$ (or to $6$), of which there are infinitely many, are not the sum of two totally square integers.

Alternatively, we may let $S(a)$ denote the sum of the digits of a number $a$ and note that $S(a)\equiv 0,1(\mathrm{mod}3)$ when $a$ is totally square. When $n=n_1+n_2$ is the sum of two totally square integers $n_1,n_2$ and if $n\equiv 0(\mathrm{mod}3)$, we have $0\equiv n\equiv S(n_1)+S(n_2)(\mathrm{mod}3)$ which implies $S(n_1)\equiv S(n_2)\equiv 0(\mathrm{mod}3)$. A square number that is divisible by $3$ actually is divisible by $9$, hence $S(n_1)\equiv S(n_2)\equiv 0(\mathrm{mod}9)$ and so $n\equiv S(n_1)+S(n_2)\equiv 0(\mathrm{mod}9)$. Therefore, positive integers that are congruent to $3$ or $6$ (mod $9$) cannot be the sum of two totally square integers, e.g. the numbers $3\cdot 10^k$ for all $k\ge 0$.