is $\displaystyle \sum_{n=1}^{p-1}{n}$ always divisible by p if p is prime? Thanks for any help
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Originally Posted by hmmmm is $\displaystyle \sum_{n=1}^{p-1}{n}$ always divisible by p if p is prime? Thanks for any help Yes, unless $\displaystyle p=2$. In fact, it is true as long as $\displaystyle p$ is ANY odd number. Try rewriting the sum (for $\displaystyle p>2$, in which case $\displaystyle p$ is odd): $\displaystyle [1+(p-1)]+[2+(p-2)]+\cdots +[\frac{p-1}{2}+\frac{p+1}{2}]$
$\displaystyle \sum_{n=1}^{p-1}{n}=\frac{p-1}{2}*p$ as p-1 is even (so divisible by 2) then it is divisible by p?
Exactly. Writing the sum as I did above, you immediately get what you have just written.
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