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Note that for $(x+1)(x+2)$ the coefficient of the linear term is divisible by $3$. Prove that in general for $$f(x)=(x+1)(x+2)(x+3)\cdots(x+p-1)$$ the coefficient of $x$ is divisible by $p$ where $p$ is an arbitrary odd prime.

Piquito
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    $\mod p$ that polynomial is just $x^{p-1}-1$ by Fermat's little theorem. – lulu Dec 13 '19 at 12:02
  • Evaluate the polynomial $f(x)-(x^{p-1}-1)$, of degree $\leq p-2$, at $x=-1,-2,...,-p+1$. It is a multiple of $p$ for all those $p-1$ values. Since $p-1$ is larger than its degree, then all its coefficients are multiples of $p$. You can see this, for example, by noting that the resulting system, with the coefficients as unknowns has the Vandermonde matrix of the values $-1,-2,...,-p+1$ as the matrix of the system. – egorovik Dec 13 '19 at 12:09
  • Another way to prove it directly can be: The coefficient is the sum of the products of all $1,2,...,p-1$ except one of them is missing in each term. In the set $1,2,...,p-1$ each element except $1$ has another element from the set that multiplied by it leaves remainder $1$ when divided by $p$. Therefore, the remainder of the product is just the same as the remainder of $1+2+...+p-1=\frac{p(p-1)}{2}$, which is $0$. – egorovik Dec 13 '19 at 12:15
  • Another way, similar to the previous one, is to compute $\frac{f'(x)}{f(x)}$ evaluated at $x=0$ in two ways. First, at $x=0$ this quotient is $\frac{c_1}{f(0)}$, where $c_1$ is the coefficient of the term of degree $1$. Then note that $\frac{f'(x)}{f(x)}=\frac{1}{x+1}+...+\frac{1}{x+p-1}$. So, at $x=0$ we get $\frac{1}{1}+...+\frac{1}{p-1}$, which, as in the previous argument, leaves the same remainder as $1+2+...+p-1=\frac{p(p-1)}{2}$. – egorovik Dec 13 '19 at 12:21
  • @Piquito Yes, note that because each element of a product like $x_1x_3...x_{p-1}$ comes with its 'inverse' modulo $p$, then the product leaves remainder $-x_2^{-1}$. So, the sum leaves the same remainder as $1+2+...+p-1$. Only the the terms are scrambled. – egorovik Dec 13 '19 at 12:24
  • @egorovik: You are right (and very good!). Thanks you. – Piquito Dec 13 '19 at 12:29
  • Let me be more precise. Each element comes with its inverse, except for $1$, which is its own inverse, and $p-1$. That is why the remainder of each product has that $-1$ sign. Strictly speaking the sum gives the same remainder as $-(1+2+...+p-1)$. But well, that is zero anyway. – egorovik Dec 13 '19 at 12:33

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Working over the finite field $\mathbb F_p$ we consider the polynomial $$g(x)=x^{p-1}-1$$ By Fermat's Little Theorem the roots of that polynomial are the non-zero elements of the field. Comparing the lead coefficients we conclude that $$g(x)=\prod_{i=1}^{p-1}(x-i)$$

Now, as $i$ runs through the non-zero residues $\pmod p$ so does $-i$ so

$$g(x)=\prod_{i=1}^{p-1}(x+i)=f(x)$$

and we are done.

lulu
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  • The coefficient in question is (where $x_i\in{1,2,3,\cdots,p-1}$) $$x_2x_3\cdots x_{p-1}+x_1x_3\cdots x_{p-1}+\cdots +x_1x_2\cdots x_{p-3}x_{p-1}+x_1x_3\cdots x_{p-2}$$ or equivalently $$\sum_{i=1}^{p-1}\prod_{j\ne i}x_j$$ – Piquito Dec 13 '19 at 12:25
  • Yes...what's your point? My argument shows that every coefficient of $f(x)$ (other than the lead term and the constant term) is divisible by $p$. Using your example, we see that $(x+1)(x+2)=x^2+3x+2\equiv x^2+2\equiv x^2-1\pmod 3$, just to illustrate the point. – lulu Dec 13 '19 at 12:28
  • You are done.Regards. – Piquito Dec 13 '19 at 12:33
  • Note that now it is very easy to solve the following problem: https://math.stackexchange.com/questions/3473865/prove-that-if-p-is-an-odd-prime-number-then-fp-binom2p-1p-1-1-is – Piquito Dec 13 '19 at 12:41
  • I don't see that...as a rule, passing from divisibility by $p$ to divisibility by $p^2$ is a challenge, though of course there are examples where that's not hard. If you have an easy solution to that problem, post it (there)! The existing solution is not what I'd call easy. – lulu Dec 13 '19 at 12:49
  • @lulu: $\binom{2p-1}{p-1}=\dfrac{(p+1)(p+2)\cdots(p+p-1)}{(p-1)!}=\dfrac{Np^2+ap}{(p-1)!}+1\Rightarrow END$ – Piquito Dec 13 '19 at 13:30
  • Ok, looks good. Like I say, I suggest posting that as a solution to the linked problem. – lulu Dec 13 '19 at 13:37
  • I'm not interested in the kind of competition that involves the accumulation of reputation points. But it does disturb me and I am against putting downvotes without justifying why. That tempts me sometimes not to participate more in maths StackExchange, a portal that I love very much for the rest. – Piquito Dec 13 '19 at 13:47
  • I'm not suggesting that you do it for the points...indeed, you can always post your solution under "community" if you don't want to participate in that sort of competition. I do that when, for example, I feel that my post is more of a synthesis of joint work with other users than an independent effort on my part. But, as I mentioned, the existing solution to that problem is not very clear and I think yours would be better. – lulu Dec 13 '19 at 13:53