Let
![p](/media/m/1/c/8/1c85c88d10b11745150467bf9935f7de.png)
be an odd prime number. For every integer
![a,](/media/m/e/2/b/e2baaa59c9d63ae97603054801408ea3.png)
define the number
![S_a = \sum^{p-1}_{j=1} \frac{a^j}{j}.](/media/m/4/f/6/4f64c16a219b83363285182d5ac0436b.png)
Let
![m,n \in \mathbb{Z},](/media/m/5/c/3/5c38ef7ecf37e4ddbffaa81d640516d8.png)
such that
![S_3 + S_4 - 3S_2 = \frac{m}{n}.](/media/m/d/5/9/d59cfc3855d36510d5ff788793768d0b.png)
Prove that
![p](/media/m/1/c/8/1c85c88d10b11745150467bf9935f7de.png)
divides
![m.](/media/m/c/6/9/c69a166e977669b9486ba58ac831eea1.png)
Proposed by Romeo Meštrović, Montenegro
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Let $p$ be an odd prime number. For every integer $a,$ define the number $S_a = \sum^{p-1}_{j=1} \frac{a^j}{j}.$ Let $m,n \in \mathbb{Z},$ such that $S_3 + S_4 - 3S_2 = \frac{m}{n}.$ Prove that $p$ divides $m.$
Proposed by Romeo Meštrović, Montenegro