SAUDI ARABIAN MATHEMATICAL COMPETITIONS

We consider the degree of polynomial $P$:

Case 1: If $\mathrm{deg}P=0$ then $P(x)\equiv c$ for some $c\in \mathbb{Z}$, the given condition becomes $S(c)=c$ which holds if and only if $1\le c\le 9$.

Case 2: If $\mathrm{deg}P=1$. We notice that $S(m+n)\le S(m)+S(n)$ for all positive integers $m,n$ and the equality occurs when there is no carry in the addition $m+n$. Let $P(x)=ax+b$ for $a,b\in \mathbb{Z}$ and $a\ne 0$. Since $S(P(n))=P(S(n))$ for $n\ge 2017$ then $a$ must be positive. The given condition becomes $S(an+b)=aS(n)+b$ for $n\ge 2017$. Putting $n=2025$ and $n=2020$, we get $$S(2025a+b)-S(2020a+b)=(aS(2025)+b)-(aS(2020)+b)=5a.$$ We also have $$S(2025a+b)=S((2020a+b)+5a)\le S(2020a+b)+S(5a).$$ These imply that $S(5a)\ge 5a$, as $a\ge 1$, this holds only when $a=1$. Then we have $S(n+1+b)-S(n+b)=S(n+1)-S(n)$.

1. If $b>0$ then choose $n$ such that $n+1+b=10^k$ for some $k$ big enough then all digits of $n+b$ are $9$ then the left hand side is $1-9k$. Also note that $n$ is a positive integer less than $10^k-1$, then $S(n)<9k$, which means $S(n+1)-S(n)\ge 1-(9k-1)=2-9k>1-9k$, a contradiction.

2. If $b<0$, similarly, we also come to a contradiction.

Thus $b=0$ and $P(x)=x$, which is trivially satisfied.

Case 3: If $\mathrm{deg}P=d\ge 2$ then the leading term of $P$ is $a_d{n}^d$ with $a_d\ne 0$, then similarly we have $a_d>0$. Choose $n=10^k-1$ for some $k\in {\mathbb{Z}}^{+}$ big enough, then $S(P(n))=P(9k)$.

Since $P(9k)$ grows approximately as $(9k{)}^d$ while $S(P(n))$ grows approximately as a constant multiple of $k$, the given equality cannot hold for sufficiently large $k$ since $d\ge 2$.

In conclusion, $P(x)=c$ for some $c\in \{1,2,3,\dots ,9\}$ or $P(x)=x$, $\forall x$. $\square$