Problem: Is there a polynomial $p(x)$ with integer coefficients such that $p(2013)=1789$ and $p(1515)=1830$?
My attempt:
After ruling out polynomials of degree 1, 2 and 3, and with further inspection, it appeared to me that such a polynomial doesn't exist.
Is my below proof accurate?
Suppose that $p$ exists, and let
$q(x)=p(x)-1789$ and $s(x)=p(x)-1830$
then, $q(2013)=0$ and $s(1515)=0$, so clearly,
$q(x)=(x-2013)f(x)$ and $s(x)=(x-1515)g(x)$,
for some polynomials $f$ and $g$ with integer coefficients.
Let $a$, $c$ and $d$ be the constant terms of $p$, $g$ and $f$, respectively, and observe that the constant terms of $s$ and $q$ are
$-1515c=a-1830 \ $ and $-2013d=a-1789$, $ \ $ respectively. $ \ \ \ \ \ \ (*)$
Thus,
$a \equiv 1830 \equiv 315 \pmod{1515}$ and $a \equiv 1789 \pmod{2013}$.
But this system of congruence doesn't have a simultaneous solution (since $ \gcd(1515, 2013)\not\mid 1789-315$).
Hence, there's no integer value of $a$ that satisfies $(*)$, which is a contradiction.
Therefore, $p$ doesn't exist.
If the above is correct, is there perhaps a more direct way of proving this?
$p(2013)-p(1515)$ is divisible by $2013-1515=498,$ but $1789-1830=-41$ is not.
It follows from the following reasoning.
Let $p(x)=a_0x^n+a_1x^{n-1}+...+a_n,$ where $a_i\in\mathbb Z$.
Thus, $$p(m)-p(k)=a_0(m^n-k^n)+a_1(m^{n-1}-k^{n-1})+...+a_{n-1}(m-k)=$$ $$=(m-k)(a_0(m^{n-1}+m^{n-2}k+...+k^{n-1})+a_1(m^{n-2}+...+k^{n-2})+...+a_{n-1}).$$