Question

consider the polynomial
\[f(x) = (x-1)(x-2)(x-3)\cdots (x-(p-1))-(x^{p-1}-1)\\=a_{p-2}x^{p-2}+a_{p-3}x^{p-3}+\cdots+a_1x+a_0\]

show that each of the coefficient, \(a_i\) is divisible by the prime \(p\)

Answer 1

I might be an idiot for saying this but I wonder if this has anything to do with fermat's little theorem

Answer 2

Yes that \(x^{p-1}-1\) piece comes from little fermat :)

Answer 3

\[a_0=(-1)(-2)(-3) \cdots (-(p-1))+1 \\ \\ \text{ so } a_0=-(p-1)!+1 \\ \text{ so is } a_0 \text{ divisible by } p? \\ \\\]

...

Answer 4

just trying to look at a_0 right now because I thought that would be easiest

Answer 5

it should be \(+(p-1)!+1\) right ?
recall wilson

Answer 6

you are right p is odd
so p-1 is even

Answer 7

p is odd for the most part*

Answer 8

hey its not super human at all, its elementary number theory.. you will like it if you try few problems
@Pikachubacca

Answer 9

yes that gives us another way to prove wilson's thm
http://mathworld.wolfram.com/WilsonsTheorem.html

Answer 10

I think I'm lost on the other coefficients.

Answer 11

not too sure if it is useful but it seems the coefficient of \(x^{p-2}\) is given by

\(\large{\begin{align}a_{p-2} &= (-1)+(-2)+(-3)+\cdots+(-(p-1)) \\~\\
&= -\dfrac{p(p-1)}{2}\\~\\
&=-\binom{p}{2}
\end{align}}\)

Answer 12

it should p-1 is even so 2|(p-1)
so p|a_(p-2)

Answer 13

right, other coefficients look a bit complicated

Answer 14

so i guess this isn't the best route :p

Answer 15

I'm going to drink some coffee and eat some cookies
I will be back later

Answer 16

okay enjoy your coffee and snacks :)

Answer 17

idk should i ruin the fun or just give a hint :O

Answer 18

first give hint maybe..

Answer 19

ok
in Z_p of order x
\(x^{p-1}-1= (x-1)(x-2)(x-3)\cdots (x-(p-1))\)

Answer 20

right, both left hand side and right hand side are congruent to \(0\mod p\) for all \(x\in\{1,2,3,\ldots,(p-1)\}\)

i don't really know how to use that hint

Answer 21

well its proof goes nice :O