nice problem ,any solutions?

bumpppppp

Goutham wrote: $(GBR 1)$ The polynomial $P(x) = a_0x^k + a_1x^{k-1} + \cdots + a_k$, where $a_0,\cdots, a_k$ are integers, is said to be divisible by an integer $m$ if $P(x)$ is a multiple of $m$ for every integral value of $x$. Show that if $P(x)$ is divisible by $m$, then $a_0 \cdot k!$ is a multiple of $m$. If $k \geq m,$ then $a_0.k! \vdots m$ Consider $k<m$ Apply Abel Interpolation, we know that exists integers $b_k,...,b_1,b_0:$ $$P(x)=b_k(x-1)(x-2)...(x-k)+b_{k-1}(x-1)(x-2)...(x-(k-1))+...+b_1(x-1)+b_0$$Choose $x=1$ then $b_0 \vdots m$ Choose $x=2$ then $b_1.1! \vdots m$ ... By induction, we can prove: $b_i.i! \vdots m$ So, $b_k.k! \vdots m,$ or $a_0.k! \vdots m$ Goutham wrote: Also prove that if $a, k,m$ are positive integers such that $ak!$ is a multiple of $m$, then a polynomial $P(x)$ with leading term $ax^k$can be found that is divisible by $m.$ Choose $P(x)=a.(x-1)...(x-k)$, then $P(x) \vdots a.k! \vdots m, \, \forall x$

what is $a \vdots b$???? and abel interpolation

PRMOisTheHardestExam wrote: what is $a \vdots b$???? and abel interpolation In some countries, $a \vdots b$ means $b \mid a$.

The question in this topic is the 1969 Longlist formulation of the same problem in 1968 Shortlist. Dan posted a beautiful solution here.