Find all polynomials $p(x) \in \mathbb{Z}[x]$ such that for all positive integers $n,$ we have that $p(n)$ is a Palindrome number. Palindrome numbers: A number $q$ written in base $10$ is called a Palindrome number, if $q$ reads the same from left to right, as it reads from right to left. For example : $121, -123321$ are Palindrome numbers, but $113$ is not a Palindrome number. (Proposed by Aditya Guha Roy (India) and Fedor Petrov (Russia))
Problem
Source: Mathlinks Contest 2020, Problem 2
Tags: algebra, polynomial, mathlinks
62861
13.07.2020 02:57
The answer is only constant palindromes, which obviously work. In fact, even if we only assume that the first and last digits of $p(x)$ are always equal, it follows $p$ is constant. To see this, suppose $p$ satisfies this condition and let $d = \deg p$; assume for a contradiction $d > 0$. WLOG assume that $p(x) > 0$ for all $x > 0$.
Claim. $p(x) \equiv 1 \pmod{10}$ for all $x$.
Proof. Since $a - b \mid p(a) - p(b)$, it's enough to show that there are $10$ consecutive values of $p$ congruent to $1$ modulo $10$. (In fact, it's enough to find $5$ such values.) To see this, note that since $p(x)$ is asymptotic to $x^d$, there are $10$ consecutive values whose decimal expansions begin with $1$. Then their decimal expansions end with $1$, so they are congruent to $1$ modulo $10$.
But similarly $p(x) \equiv 2 \pmod{10}$ for all $x$, contradiction.
Physicsknight
13.07.2020 05:43
If $p(x)$ is non-constant,$ p(10x)$ has to be constant $(\mod 10),$ so $p(10x)$ ends in a constant digit $d$ for sufficiently large $x.$ However, the first digit of $p(10x)$ has to be a digit other than $d$ infinitely many times, since polynomials do not increase fast enough to skip over the numbers with first digit other than $d $ (this could be made a bit more rigorous by considering the $\mathrm {limit}_{x\mapsto\infty}\frac {p(10x+10)}{p(10x)} $ So $p(10x)$ is not always a palindrome if $p(x)$ is non-constant. So the only polynomials that satisfy the given conditions are constant polynomials $p(x) = c,$ where $c$ is itself a palindrome.
Physicsknight
14.07.2020 10:39
Another solution Claim- There is no non-constant $p(x)\in\mathbb Z[x] $ such that $\forall\,n\in\mathbb N.$ First and last digit of $p(n) $ is same. Proof- Assume that there is a polynomial with the above property. $P(x)=\sum_{i=0}^n a_i x_i\,\left(n\in\mathbb N*\, ,a_n\ne 0\right) $ $\mathrm {WLOG},$ let $a_n>0$ There exists a number $N_0\in\mathbb N$ . $(\star) \, \left|\sum_{i=0}^{n-1}a_i 20^{N_0}\right|< 20^{N_0 n-1} $ $(\bullet)\,\sum_{i=0}^{n-1} a_i 20^{N_0 i} $ will be the same sign $\forall\, m\ge N_0$ Consider the $2$ cases $\sum_{i=0}^{n-1} a_i 20^{m_i}\ge 0\,\forall\, m\ge N_0$ Assume that $a_n 2^m $ has $k_m $ digits. Because $(\star) $ $p(20^m) $ and $a_n\cdot 2^m $ have the same first $k_m $ digits$\,\implies a_n\cdot 2^m $ will have the first digit $=a_0^{'}\text {s} $ last digit $\,\forall\,m\in N_0.$ But $a_n\cdot 2^m $ and $a_n\cdot 2^{m+1} $ always have different first digit $\implies\text {contradiction}.$ Note that, $\sum_{i=0}^{n-1}a_i\cdot 20^{m_i}< 0\,\forall\, m\ge N_0$ Assume that $a_n\cdot 2^m-1$ have $k_m $ digits. Because of $(\bullet)\,P(20^m) $ and $a_n\cdot 2^{m}-1$ have the same first $k_m $ digits. $a_n\cdot 2^m-1$ will have the first digit $=a_0^{'}\text {s} $ last digit $\,\forall\, m\ge N_0$ But $a_n\cdot 2^m-1$ and $a_n\cdot 2^{m+1}-1$ always have different first digit.
adityaguharoy
16.07.2020 20:01
dangerousliri wrote: Find all polynomials $p(x) \in \mathbb{Z}[x]$ such that for all positive integers $n,$ we have that $p(n)$ is a Palindrome number. Palindrome numbers: A number $q$ written in base $10$ is called a Palindrome number, if $q$ reads the same from left to right, as it reads from right to left. For example : $121, -123321$ are Palindrome numbers, but $113$ is not a Palindrome number. (Proposed by Aditya Guha Roy (India) and Fedor Petrov (Russia))
We prove that such polynomials are only palyndromic constants. Assume that $\deg p=n>0$. Choose a positive integer $T$ such that $q(x)=p(x+T)$ has positive coefficients. Write $q(x)=a_0(T)+a_1(T)x+a_2(T)x^2+\ldots+a_n(T)x^n$, where $a_i(T)=p^{(i)}(T)/i!$. In particular, $a_n(T)$ does not depend on $T$, and $a_0=p(T)$. Since $q(10^M)=p(10^M+T)$ for large $M$ is a palyndrome, we see that $p(T)=a_0(T)=b\cdot 10^{r(T)}$, where $b$ is fixed (depending on $a_n$ only). This is clearly not possible, for example because $p(T)<p(T+1)<2p(T)$ for large $T$.
What can you say about the limit of the ratio of two consecutive Palindrome numbers ?
That was where the problem originated.