# Sum of powers factorization

• Aug 21st 2009, 01:57 AM
gdmath
Sum of powers factorization
Hi everybody,

I think that:

$\displaystyle N=\sum_{i=0}^{s}p^i=p^s+...+1 \quad \quad s>1$

Is factorized in prime numbers (according theorem of arithmetic), that are square free.

In other words its dinstict primes (if multiplied) give as result number N.

i.e.
$\displaystyle N=p_1^{a_1}\cdot...\cdot p_k^{a_k}$

where
$\displaystyle {a_1}=...={a_k}=1$

Is this correct?

I think this can be handled with Congruence of Clausen and von Staudt proof. Unfortunatelly i cannot find any link (with proof) into internet.

Thank you a lot
• Aug 21st 2009, 07:42 AM
Bruno J.
You mean that $\displaystyle 1+p+...+p^s=\frac{p^{s+1}-1}{p-1}$ is always squarefree?

Or does $\displaystyle p$ have to be prime?
• Aug 21st 2009, 07:44 AM
Bruno J.
In any case a counter-example is $\displaystyle 1+7+7^2+7^3=2^45^2$
• Aug 21st 2009, 09:03 AM
gdmath
Ok counter example is definite thanks
• Aug 21st 2009, 09:37 AM
Bruno J.
I suggest always checking a couple of cases before conjecturing something, let alone trying to prove it. Counter-examples to your conjecture abound. It took me about three tries to find one.