Prove that for every p - prime, $p>3$ the numerator of sum of $p-1$ summands of harmonic series is divisible by $p$ square.

Asked Feb 16 '20 at 19:11

Active Feb 16 '20 at 19:11

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I've written 1 A4 but so far without victory. This can't require advanced tools, as it is an exercise after chapter with only Euler's Theorem, Wilson's Theorem and CRT.

I've tried with Wilson's by multiplying everything by $(p-1)!$ and then write $(p-1)!=(p-1)*(p-2)!$ and later calculus - but it does not work so far.

Thanks in advance for help.

asked Feb 16 '20 at 19:11

robin3210

2

This is known as Wolstemholme's theorem. – lulu Feb 16 '20 at 19:22
Note: in that duplicate the problem statement omits the (necessary requirement $p>3$) but the answer does state (and use) this assumption. – lulu Feb 16 '20 at 19:27
Yes, it answers my question. Thank You very much Mr @lulu – robin3210 Feb 16 '20 at 19:30

Prove that for every p - prime, $p>3$ the numerator of sum of $p-1$ summands of harmonic series is divisible by $p$ square.

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