Question

Establish below :-

\(1^n+2^n+3^n + ... + (p-1)^n \equiv 0 (\mod{p}) ~~if~~ (p-1) \not{|} n\)

\(1^n+2^n+3^n + ... + (p-1)^n \equiv -1 (\mod{p}) ~~if ~~ (p-1) | n\)

using basic properties of congruences + Fermat's little theorem. \(p\) is a prime > 2

Answer 1

You stole my question! @ganeshie8

Answer 2

yahh its an interesting one, worth spending time :)

Answer 3

Wanna borrow my textbook while you're at it? :)

Answer 4

sure, im still working on it xD

Answer 5

Use \mbox to make \(\mbox{if}\) :)

Answer 6

NB: I skipped heck-a-lot of steps
\[\mbox{if }(p-1)|n,\]\[\large\begin{array}{clrr}
&1^n+2^n+3^n+4^n+...+(p-1)^n&\\
=&1^{m(p-1)}+2^{m(p-1)}+...+(p-1)^{m(p-1)}\\
\equiv&\begin{array}{c}\underbrace{1+1+1+...+1}\\p-1\end{array}&\mod p&\mbox{(FLT)}\\
=&p-1\\
\equiv&-1&\mod p&\\
\end{array}\]

Answer 7

looks great ! part 1 looks tough

Answer 8

this problem can be solved using order of an element \[ord_pg=n\]
where \[\phi(p)=p-1|n\]
\[g^n\equiv1\mod p\\ 1\le g<p\]

if
\[\phi(p)=p-1\cancel{|}n\]
then \[g^n\equiv0\mod p\]

Answer 9

since the is no least element to make gaurantee fermat

Answer 10

primitive roots can also give a better intuition here