1. ## help

can anyone help me with these 2 super hard questions...

1. If n is positive integer what is the number of solutions (x,y) (with x and y postive integers) to the equation
1/x +1/y =1/n

2. let p be an odd prime, and i be an integer, 0<= i <= p-2. show that 1^i +2^i +3^i +.....+ (p-1)^i = 0 (mod p)

can anyone help me with these 2 super hard questions...

1. If n is positive integer what is the number of solutions (x,y) (with x and y positive integers) to the equation
1/x +1/y =1/n

2. let p be an odd prime, and i be an integer, 0<= i <= p-2. show that 1^i +2^i +3^i +.....+ (p-1)^i = 0 (mod p)
1) is not too hard. If you multiply out the fractions, then the equation $\displaystyle \tfrac1x+\tfrac1y = \tfrac1n$ can be written $\displaystyle (x-n)(y-n)=n^2$. So the number of solutions is the same as the number of ways of factorising $\displaystyle n^2$ as a product of two factors (which I'll leave you to think about).

2) seems to be harder (unless I'm missing a simple solution). First, the question as stated is wrong, since the result is clearly false when i=0. So we want to prove it for $\displaystyle 1\leqslant i \leqslant p-2$.

The only way I can see to do this is to quote the result that the multiplicative group $\displaystyle \mathbb{Z}_p^*$ of nonzero elements of $\displaystyle \mathbb{Z}_p$ is cyclic. Let r be a generator of this group. Then $\displaystyle \sum_{k=1}^{p-1}k^i = \sum_{j=1}^{p-1}r^{ji}$, because the elements $\displaystyle r,r^2,\ldots,r^{p-1}$ are just 1,2,...,p-1 in some order. But if $\displaystyle 1\leqslant i \leqslant p-2$ then the map consisting of multiplication by i is a bijection of $\displaystyle \mathbb{Z}_p^*$. So the terms $\displaystyle r^i,r^{2i},\ldots,r^{(p-1)i}$ are also just 1,2,...,p-1 in some order. That reduces the problem to showing that $\displaystyle \sum_{k=1}^{p-1}k\equiv0\!\!\!\pmod p$. But that is obviously true because $\displaystyle \sum_{k=1}^{p-1}k = \tfrac12p(p-1)$.

3. Originally Posted by Opalg
(unless I'm missing a simple solution).
That is the standard (simple) way of proving it.
Except I just write out $\displaystyle 1^i + ... + (p-1)^i \equiv r^i + r^{2i} + ... + r^{(p-1)i}$
And now apply geometric series.
We from there that $\displaystyle 1^i + ... + (p-1)^i \equiv 0$ if $\displaystyle i \not | (p-1)$.