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ganeshie8
 one year ago
consider the polynomial
\[f(x) = (x1)(x2)(x3)\cdots (x(p1))(x^{p1}1)\\=a_{p2}x^{p2}+a_{p3}x^{p3}+\cdots+a_1x+a_0\]
show that each of the coefficient, \(a_i\) is divisible by the prime \(p\)
ganeshie8
 one year ago
consider the polynomial \[f(x) = (x1)(x2)(x3)\cdots (x(p1))(x^{p1}1)\\=a_{p2}x^{p2}+a_{p3}x^{p3}+\cdots+a_1x+a_0\] show that each of the coefficient, \(a_i\) is divisible by the prime \(p\)

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freckles
 one year ago
Best ResponseYou've already chosen the best response.2I might be an idiot for saying this but I wonder if this has anything to do with fermat's little theorem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yes that \(x^{p1}1\) piece comes from little fermat :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[a_0=(1)(2)(3) \cdots ((p1))+1 \\ \\ \text{ so } a_0=(p1)!+1 \\ \text{ so is } a_0 \text{ divisible by } p? \\ \\\] ...

freckles
 one year ago
Best ResponseYou've already chosen the best response.2just trying to look at a_0 right now because I thought that would be easiest

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0it should be \(+(p1)!+1\) right ? recall wilson

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you are right p is odd so p1 is even

freckles
 one year ago
Best ResponseYou've already chosen the best response.2p is odd for the most part*

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0hey its not super human at all, its elementary number theory.. you will like it if you try few problems @Pikachubacca

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0yes that gives us another way to prove wilson's thm http://mathworld.wolfram.com/WilsonsTheorem.html

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I think I'm lost on the other coefficients.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0not too sure if it is useful but it seems the coefficient of \(x^{p2}\) is given by \(\large{\begin{align}a_{p2} &= (1)+(2)+(3)+\cdots+((p1)) \\~\\ &= \dfrac{p(p1)}{2}\\~\\ &=\binom{p}{2} \end{align}}\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2it should p1 is even so 2(p1) so pa_(p2)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0right, other coefficients look a bit complicated

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so i guess this isn't the best route :p

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I'm going to drink some coffee and eat some cookies I will be back later

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0okay enjoy your coffee and snacks :)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2idk should i ruin the fun or just give a hint :O

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0first give hint maybe..

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2ok in Z_p of order x \(x^{p1}1= (x1)(x2)(x3)\cdots (x(p1))\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0right, both left hand side and right hand side are congruent to \(0\mod p\) for all \(x\in\{1,2,3,\ldots,(p1)\}\) i don't really know how to use that hint

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2well its proof goes nice :O
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