nice problem

Solved with upwgs_2008 (55th post btw) FIRST POST OF the new year yay By taking $a_1 \mid a_0$, we get that $f(a) \mid f(b)$ if $a \mid b$. Now, by taking $a_0 = 1$, $a_1 = 2$, $a_2 = n+1$, $a_{2k+1} = n$, note that we can make $k$ arbitrarily, we get the absolute value of $a_{2k+1}x^{2k+1}$ will become arbitrarily large, but the absolute value of $a_2x^2+a_1x+a_0$ is fixed. Thus, we must have $x = -1$ is the root, and therefore, $f(n+1)-f(n) = f(2)-f(1)$. So, $f$ is an arithmetic progression. And, since $f(1) \mid f(2)$, we have that the first term divides the common difference. Let both of them be $a$ and $ka$ respectively. Then, $f(2)$ divides $f(4)$, so $k+1 \mid 3k+1$, so $k+1 \mid 2$, so $k = 1$. Therefore, $f(x) = f(1)x$. This is the only solution which clearly works @below here is a fix As we make $k$ arbitrarily large, we keep inserting $1$s in between. Since $f(1) \mid f(n)$, and since any root is negative, if it is not $-1$ then, the absolute value of $f(1)+f(n+1)x+\dots+f(1)x^{2k-1}$smaller than $f(n)x^{2k}$, so the root is $-1$ and we again get the required relationship

@above Hate to be that person, but your proof is incorrect. You in fact cannot make $k$ arbitrarily large, since no zero coefficients are allowed in the polynomials. Each $a_i$ must be a positive integer, but you seem to be ignoring those coefficients.

Sol:- $a_0|a_1 \iff a_0+a_1x$ has integer root $\implies f(a_0)+f(a_1)x$ has integer root $\iff f(a_0)|f(a_1)$.So $a_0|a_1 \implies f(a_0)|f(a_1)$. $kx^2+2kx+k=k(x^2+2x+1)$ has integer root $\implies f(k)x^2+f(2k)x+f(k)$ has integer root $\iff x^2+\frac{f(2k)}{f(k)}x+1$ has integer root. Since $k|2k \implies \frac{f(2k)}{f(k)}$ is positive integer. This forces $\frac{f(2k)}{f(k)}=2$. In particular we have $f(2)=2f(1)$ and $f(4)=4f(1)$. $x^2+3x+2=(x+2)(x+1)$ has integer roots $\implies f(1)x^2+f(3)x+f(2)$ has integer root $\implies x^2+\frac{f(3)}{f(1)}x+2$ has integer root and since $\frac{f(3)}{f(1)}$ is a positive integer ,this forces $\frac{f(3)}{f(1)}=3 \implies f(3)=3f(1)$. Now we prove by strong induction that $f(n)=nf(1)$.We have already proved the base cases so assume $n \geq 5$. Suppose that we have established that $f(k)=kf(1)$ for all $ k \in \{1,...,n-1\}$. Consider the quadratic equations $(x+n-1)(x+1)=x^2+nx+n-1$ and $(x+n-2)(x+2)=x^2+nx+2n-2$ , both have integral roots. So $f(1)x^2+f(n)x+f(n-1)$ and $f(1)x^2+f(n)x+f(2n-2)$ have integral root. We know $\frac{f(n-1)}{f(1)}=n-1$ is true by inductive hypothesis and $2=\frac{f(2k)}{f(k)} \implies \frac{f(2n-2)}{f(1)}=2\frac{f(n-1)}{f(1)}=2n-2$. So it follows that $x^2+\frac{f(n)}{f(1)}x+n-1$ and $x^2+\frac{f(n)}{f(1)}+2n-2$ have integral root. Since $\frac{f(n)}{f(1)}$ is a positive integer, this forces $\frac{f(n)}{f(1)}=n \implies f(n)=nf(1)$. This completes the induction. Substituting this in the original expression we have that $f(a_0)+f(a_1)x+...+f(a_n)x^n=f(1)(a_0+a_1x+...+a_nx^n)$. Since the $2$ polynomials share same set of roots so when one of them has integral root so does the other. Hence any $f(x)=cx$ works where $c$ is a natural number.

I claim the answer is only $f(x) = kx$. It is trivial to see that all of these functions work. Observe that $a_0 = q, a_1 = p$ with $p | q$ forces $f(q) | f(p)$. Now $f(1)$ divides all $f(i)$, and scaling doesn't change the prescence of any integer roots, so WLOG we find all solutions with $f(1) = 1$ and use this to find the rest of the solutions. Now we (strong) inductively show $f(x) = x$ for each integer $m$.lI We also assign a set of special "determining" polynomials for each of the primes. The determining polynomial for $2$ is $x^2 + 2x + 1$, meaning that if we know $f(n) = n$, we can easily show $f(2n) = 2n$ using $nx^2 + 2nx +n$ having an integer root, we must have $nx^2 + f(2n)x + n $ has an integer root, implying $f(2n)^2 - 4n^2$ is a square, since $n | f(2n)$ we have $k^2n^2 - 4n^2$ is a square, forcing $k = 4$. Case 1: $m$ is prime. Consider the polynomial given by $x^2 + mx + m - 1$, then we have $x^2+ f(m)x + m - 1 $ has an integer root, implying $f(m)^2 - 4(m - 1)$ is a perfect square, meaning that $f(m)$ is the sum of two numbers that multiply to $m - 1$ (trivial by difference of squares). This implies that $f(m) \ge m$ Now we show $f(m + 1) = m + 1$, since $m + 1$ is even and we know $f(\frac{m+1}{2})$ we can use the proof highlighted above. Now consider the polynomial given by $x^2 + (m + 1)x + m$, since it has integer roots we are forced that $(m + 1)^2 - 4f(m)$ is an integer, meaning that two divisors of $f(m)$ sum to $m + 1$ by difference of squares. This then implies that $f(m) \le m$, forcing $f(m) = m$. Assign the determining polynomials of $m$ as the polynomials $x^2 + mx + m - 1$, $x^2 + (m + 1)x + m$. Case 2: $m$ is not prime. If $m$ is even, we can use the argument above and we are done.Otherwise, consider the determining polynomials for some prime factor $p$ of $m$, repeating the proof for $p$ with the determining polynomials all having coefficients multiplied by $\frac{m}{p}$ yields the desired results. Note that in the first polynomial, we know the outputs of all coefficients except $\frac mp$, in the second polynomial we do not know the output of $m + \frac{m}{p}$, but since it is odd and $\frac mp < m$ we do know the output of half of it, so repeating the procedure given above allows us to calculate its output, thus we know exactly the same givens as before and can repeat the exact same difference of squares bounding argument, just scaled.

Beautiful problem, really enjoyed solving it.