ganeshie8
  • ganeshie8
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 \(p\)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
I might be an idiot for saying this but I wonder if this has anything to do with fermat's little theorem
ganeshie8
  • ganeshie8
Yes that \(x^{p-1}-1\) piece comes from little fermat :)
freckles
  • freckles
\[a_0=(-1)(-2)(-3) \cdots (-(p-1))+1 \\ \\ \text{ so } a_0=-(p-1)!+1 \\ \text{ so is } a_0 \text{ divisible by } p? \\ \\\] ...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

freckles
  • freckles
just trying to look at a_0 right now because I thought that would be easiest
ganeshie8
  • ganeshie8
it should be \(+(p-1)!+1\) right ? recall wilson
freckles
  • freckles
you are right p is odd so p-1 is even
freckles
  • freckles
p is odd for the most part*
ganeshie8
  • ganeshie8
hey its not super human at all, its elementary number theory.. you will like it if you try few problems @Pikachubacca
ganeshie8
  • ganeshie8
yes that gives us another way to prove wilson's thm http://mathworld.wolfram.com/WilsonsTheorem.html
freckles
  • freckles
I think I'm lost on the other coefficients.
ganeshie8
  • ganeshie8
not too sure if it is useful but it seems the coefficient of \(x^{p-2}\) is given by \(\large{\begin{align}a_{p-2} &= (-1)+(-2)+(-3)+\cdots+(-(p-1)) \\~\\ &= -\dfrac{p(p-1)}{2}\\~\\ &=-\binom{p}{2} \end{align}}\)
freckles
  • freckles
it should p-1 is even so 2|(p-1) so p|a_(p-2)
ganeshie8
  • ganeshie8
right, other coefficients look a bit complicated
freckles
  • freckles
so i guess this isn't the best route :p
freckles
  • freckles
I'm going to drink some coffee and eat some cookies I will be back later
ganeshie8
  • ganeshie8
okay enjoy your coffee and snacks :)
ikram002p
  • ikram002p
idk should i ruin the fun or just give a hint :O
ganeshie8
  • ganeshie8
first give hint maybe..
ikram002p
  • ikram002p
ok in Z_p of order x \(x^{p-1}-1= (x-1)(x-2)(x-3)\cdots (x-(p-1))\)
ganeshie8
  • ganeshie8
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
ikram002p
  • ikram002p
well its proof goes nice :O

Looking for something else?

Not the answer you are looking for? Search for more explanations.