@above You just showed that if $P=\sum a_n x^n$ and $Q=\sum b_n x^n$ then $a_n=10^k b_n$ for all $n>0$, where $k$ might depend on $n$. How does this finish?

Phorphyrion wrote: @above You just showed that if $P=\sum a_n x^n$ and $Q=\sum b_n x^n$ then $a_n=10^k b_n$ for all $n>0$, where $k$ might depend on $n$. How does this finish? I think we can resolve this issue by some-sort of induction, I haven't worked out the details though, I will try to fix the solution, Thanks. Fixed now!

This problem is proposed by Yahya Motevassel from Iran. Lemma : If for some natural numbers $ a , b $ we have that for any number $n \in \mathbb{N}$ : $$S(an)=S(bn)$$Then there exist an integer $k$ such that $a=10^{k}b$ Proof : Fist note we can suppose that $10$ doesn't divide $a , b$ and if $a > b$ , then $a<10b$. Suppose otherwise and let $a>10b$ , then for each natural number $n$ the number of decimal digits of $bn$ is less than $an$ and if we determine a number $n \in \mathbb{N}$ such that the number of $an$ digits equals to $k$ and the last digit of $an$ equals to $9$ ( obviously we can pick such $n$ ) , then while the first digit of $a$ is not zero , we can get that : $$S(a(n+10^{k-1})) \le S(a)+S(an)-9$$On the other hand , since number of digits of $bn$ is less than $k$ , we have $S(b(n+10^{k-1}))=S(b)+S(bn)$ which is a contradiction while $S(a(n+10^{k-1}))$ and $S(b(n+10^{k-1}))$ must be equal. Now if we have $gcd(a , 10)=1$ , then by putting $n=\frac{10^{\varphi(a)}-1}{a}$ we gat that $S(an)=S(bn)=9\varphi(a)$ which since $bn$ has at most $\varphi(a)$ decimal digits and is less that $10^{\varphi(a)}-1$ , is a contradiction. So suppose that for some natural number $t$ , we have $v_{2}(a)=t$ and $a=2^ta'$. ( We can check the case that $5|a$ similarly ) Now for each number $k \equiv t \pmod {\varphi(a')}$ we define $n_k=\frac{10^{k}-10^{t}}{a}$ , and since $S(an_k)=9(n_{k}-t)$ , there exist non-negative integers $s_1 < s_2 < ... < s_m$ and digits $1 \le r_i \le 9$ such that $\sum{r_i} =9t$ and we have : $$bn_k=b(\frac{10^{k}-10^{t}}{a})=(10^{k}-1)-(10^{s_m}r_m + 10^{s_{m-1}}r_{m-1} + ... + 10^{s_1}r_1) \implies$$$$10^{k-t}(10^{t}a'-5^{t}b)+5^{t}b-a'=10^{s_m}r_{m}a'+...+10^{s_1}r_{1}a' ( I )$$So while $\sum{r_i} =9t$ and values of $r_i$ are bounded , by pigeonhole principle for infinity many proper numbers $k$ , values of $r_1 , r_2 , ... , r_m$ are constant. Now by considering large enough numbers $k$ and checking identity $(I)$ modulo $10$ , since $gcd(a' , 10)=1$ we have $s_1=0$. Again , if the value of $5^{t}b-a'-10^{s_1}r_{1}a'$ is non-zero , for large enough $k$ the value of $s_2$ must be equal to $v_{10}(5^{t}b-a'-10^{s_1}r_{1}a')$ and by continuing this process , while $10^{t}a' \neq 5^{t}b$ and left side of identity $I$ grows large and $m$ is fixed , for some number $i \le m$ we have : $$5^{t}b-a'=10^{s_i}r_{i}a'+...+10^{s_1}r_{1}a' \implies a' | 5^{t}b \implies a' | b$$Thus for a number $k \in \mathbb{N}$ , we have $b=ka'$ and if we put $5^{t}n$ in equality $S(2^{t}a'n)=S(ka'n)$ and also if $v_2(k)=s$ , one can see that : $$S(a'n)=S(5^{t-s}k'a'n)$$Which $s<t$ ( $a>b$ implies $2^t>k$ ) and $a' , k'$ are coprime to $10$. So while we know that ratio of these numbers must be less than $10$ , we get $t=s+1$ and $k'=1$ which by putting $2n$ instead of $n$ , means : $$S(a'n)=S(5a'n) \implies S(a'n)=S(2a'n)$$Now determine natural numbers $n , m$ in such a way that $a'm \equiv 4 \pmod {10}$ and $a'n$ be a $k$-digit number that $10.10^{k-1}<a'n<15.10^{k-1}$. So by putting $10^{k-1}m+n$ instead of $n$ , note that the first digit of $a'm$ equals to $4$ and the last digit of $a'n$ equals to $1$ , we have : $$S(a'(10^{k-1}m+n))=S(a'm)+S(a'n)$$And also note that the first digit of $2a'm$ equals to $8$ and the last digit of $2a'n$ equals to $2$ , we can get : $$S(2a'(10^{k-1}m+n)) \le S(2a'm)+S(2a'n)-9=S(a'm)+S(a'n)-9$$Which is a contradiction and we're done. Now suppose that $P(x)=a_nx^n+...+a_{1}x+a_0$ and $Q(x)=b_mx^m+...+b_{1}x+b_0$. So for each number $x \in \mathbb{N}$ , by taking large enough natural $k$ one can see that : $$S(P(10^{k}x))=S(Q(10^{k}x)) \implies \sum{S(a_{i}x^{i})}=\sum{S(b_{j}x^{j})} ( II )$$Again with same method , in identity $(II)$ and for large enough natural $t$ we get that : $$S(a_{i}(x+10^t)^{i})=S(a_ix^i+ia_ix^{i-1}(10^t)+...+a_i(10^{ti}))=\sum{S(\binom{i}{j}a_{i}x^{i-j})} \implies$$$$(S(a_{n}x^{n})+...+S(a_1x)+S(a_0))+(S(na_nx^{n-1})+S((n-1)a_{n-1}x^{n-2})+...+S(a_1))+...=(S(b_{m}x^{m})+...+S(b_1x)+S(b_0))+(S(mb_mx^{m-1})+S((m-1)b_{m-1}x^{m-2})+...+S(b_1))+...$$Thus in this new identity , maximum degrees of monomials in left and right hand side decrease by one. ( $na_nx^{n-1}$ and $mb_mx^{m-1}$ ) And by continuing this process , putting $x+10^t$ and decreasing maximum degrees by one , obviously $deg(P)=deg(Q)$ and there exist an integer $c$ such that : $$S(n!a_nx)=S(n!b_nx)+c$$And by taking large natural $t$ we have : $$S(n!a_n(x+10^t))=S(n!a_nx)+S(n!a_n)=S(n!b_nx)+S(n!b_n)+2c=S(n!b_n(x+10^t))+c=S(n!b_nx)+S(n!b_n)+c \implies c=0$$So by lemma , there exist an integer $k_n$ such that $a_n = 10^{k_n}b_n$ and from identity $(II)$ we get : $$\sum_{i \le n-1}{S(a_{i}x^{i})}=\sum_{i \le n-1}{S(b_{i}x^{i})}$$And with the same way , for some $k_{n-1} \in \mathbb{Z}$ , $a_{n-1}=10^{k_{n-1}}b_{n-1}$ and values $k_1 , ... , k_{n-2}$ will define as well. So for each two polynomials that always hold $S(P(x))=S(Q(x))$ , ratio of their corresponding coefficients is an integer power of $10$. Now suppose that for a non negative integer $k$ , $a_n = 10^{k}b_n$ and for each natural number $t$ , consider polynomials $P(x+t)$ and $10^{k}Q(x+t)$. ( As $P_0$ and $Q_0$ ). Now while always we have $S(P_{0}(x))=S(Q_{0}(x))$ , ratio of their corresponding coefficients is always an integer power of $10$ and by checking coefficient of $x^{n-1}$ in these polynomials we have : $$\frac{na_{n}t+a_{n-1}}{na_{n}t+10^{k}b_{n-1}}=10^{k_t}$$Note that for each large enough $t$ we have : $$10(na_{n}t+10^{k}b_{n-1}) > na_{n}t+a_{n-1} , 10(na_{n}t+a_{n-1}) > na_{n}t+10^{k}b_{n-1} \implies k_t=0 \implies a_{n-1}=10^{k}b_{n-1}$$And again by checking coefficient of $x^{n-2}$ : $$\frac{\binom{n}{2}a_{n}t^2+(n-1)a_{n-1}t+a_{n-2}}{\binom{n}{2}a_{n}t^2+(n-1)a_{n-1}t+10^{k}b_{n-2}}=10^{k'_t}$$Which similarly implies that $a_{n-2}=10^{k}b_{n-2}$ and also similarly for each $1 \le i \le n$ we have $a_{i}=10^{k}b_{i}$ and we're done !

By using a carry collision method similar to 2022 A7 and differential induction of degree, it can be transformed into a case when degP=degQ=1. Then, by using cyclic decimals and Euler's theorem, this problem can be solved. It's really a difficult problem!

Here's another way to prove that if $\forall n: P(an)=P(bn) \rightarrow \exists k: a = 10^kb$. Note that we can assume $10$ doesn't divide $a$ or $b$. and if $a\ge b$, then $a\le10b$. Suppose otherwise and let $a>10b$ , then for each natural number $n$ the number of decimal digits of $bn$ is less than $an$ and if we determine a number $n \in \mathbb{N}$ such that the number of $an$ digits equals to $k$ and the last digit of $an$ equals to $9$ ( obviously we can pick such $n$ ) , then while the first digit of $a$ is not zero , we can get that : $$S(a(n+10^{k-1})) \le S(a)+S(an)-9$$On the other hand , since number of digits of $bn$ is less than $k$ , we have $S(b(n+10^{k-1}))=S(b)+S(bn)$ which is a contradiction while $S(a(n+10^{k-1}))$ and $S(b(n+10^{k-1}))$ must be equal. We'll prove that $\forall t: S(a^tn) = S(b^tn)$. This follows from noticing that $S(a^tn) = s(a^{t-1}b^1 n) = \cdots S(a^m b^{t-m}n) = \cdots = S(b^t n)$ To finish notice we must have $\forall t: (\frac{a}{b})^t \le 10$ and if $a>b$, $(\frac{a}{b})^t$ gets arbitrary large which is a contradiction and so $a=b$ and we're done. The rest is the same as @2above.