4

Let $p,q$ are prime numbers and $n$ is a even number such that : $p^n+p^{n-1}+...+p+1=q^2+q+1$

Find $p,q$?

I think :

$p^n+p^{n-1}+...+p+1=q^2+q+1\Rightarrow p^n+p^{n-1}+...+p=q(q+1)\Rightarrow p|q(q+1)\Rightarrow p|q$ or $p|q+1$

If $p|q$ then $p=q$. I can solve this

If $p|q+1$, I don't know how to do next !?

Lê Tấn Khang
  • 1,113

1 Answers1

0

Partial answer: Suppose $p\neq q$. $q\neq 2$ and $q$ doesn't divide $n$. Assume

$$p^{n}+\cdots +p+1=q^2+q+1$$

with $n$ an even number. Using geometric sum, we have

$$\frac{p^{n+1}-1}{p-1}=q^2+q+1$$

and therefore

$$p^{n+1}-1=(p-1)(q^2+q+1)$$

If we take modulo $q$, we obtain

$$p^{n+1}-1\equiv p-1 \mod q$$

and thus, since $p\neq q$

$$p^n\equiv 1\mod q$$

Let $m$ be the multiplicative order of $p$ in the group $\mathbb{Z_q}$, then $m$ divides $n$ and $q$ (Lagrange), hence, since $n$ and $q$ are relatively prime $m=1$. This implies

$$p\equiv 1\mod q\ .$$

At this point, we return to the original equality

$$p^{n}+\cdots +p+1=q^2+q+1$$

If we take modulo $q$ at both sides, we get

$$1+1+\cdots+1\equiv 1 \mod q$$

where we sum $1$ $n+1$ times, i.e.

$$n+1\equiv 1\mod q$$

and so $n\equiv 0\mod q$, but this is absurd since $q$ doesn't divide $n$.

Daniel
  • 6,999
  • 3
    I've done a big mistake. I'm sorry. I thought the order of the multiplicative group $\mathbb{Z_q}$ is $q$, but no, it's $q-1$, I'll see if I can fix it (it will work if we make the aditional assumption that $n$ and $q-1$ are relatively primes, but they're not since $2$ divides both). – Daniel Apr 06 '15 at 15:38