Let $p$ prime be a number and let $a$ be a natural number so that $a<p$. prove that $a|p+1$ if and only if there are $M,N$ that are naturals so that $a/p = 1/N + 1/M$. I have tried a lot of algebraic games to get to something like $p+1 = a\pmod{\text{something}}$ to show that $a|p+1$ but I had no success. I really don't know where to begin.
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Is it $M=m$ and $N=n$? – PQH Nov 30 '20 at 18:50
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yes, sorry for the mistake I have edited it. – itamar Nov 30 '20 at 18:52
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Then, first try $p+1=a\cdot s$ and divide by $ps$. – PQH Nov 30 '20 at 18:54
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thank you i will try that! I'll come back if I couldn't get it. but will it help for both sides of the statement? – itamar Nov 30 '20 at 18:56
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That helps with the easy direction. – PQH Nov 30 '20 at 18:59
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and the other direction? do you have any clue? – itamar Nov 30 '20 at 19:00
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You will need that $p$ is a prime. – PQH Nov 30 '20 at 19:02
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The gives you one direction. For the other, do something similar. Show that at least one of n, m is a multiple of $p$, then show that $n = pm$. – Calvin Lin Nov 30 '20 at 19:02
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(For the harder direction.)
Hint: Show that $ \frac{a}{p} = \frac{ n+m}{nm} $ implies that $p \mid nm$.
WLOG, let $ m = pk$.
Hint: Show that $a = \frac{ n + pk}{nk} $ implies that $ k \mid n$ and $ n \mid pk$.
Conclude that $ n =k $ or $ n = pk$.
Hence, either $ \frac{a}{p} = \frac{1}{k} + \frac{1}{pk} = \frac{ p+1}{pk}$ or $ \frac{a}{p} = \frac{1}{pk} + \frac{1}{pk} = \frac{2}{pk}$.
So either $ ak = p+1$ or $ a = 1, 2$, and we are done.
Calvin Lin
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