Thread: Sum of reciprocals of phi(p)

1. Sum of reciprocals of phi(p)

Write $\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{p-1} = \frac{a}{b}$, with $gcd(a,b)=1$. Show that $p^2$ divides a if $p \geq 5$

I have no clue how to express it as $\frac{a}{b}$.

2. Originally Posted by comssa
Write $\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{p-1} = \frac{a}{b}$, with $gcd(a,b)=1$. Show that $p^2$ divides a if $p \geq 5$

I see that $\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{p-1} = \frac{1}{\phi(2)} + \frac{1}{\phi(3)} + ... + \frac{1}{\phi(p)}$
but I have no clue how to express it as $\frac{a}{b}$.

I can't understand what you think you can see here: $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{p-1}\neq \frac{1}{\phi(2)}+\frac{1}{\phi(3)}+\frac{1}{\phi( 4)}+...+\frac{1}{\phi(p)}$ , since, for example, $\phi(4)=2\neq 3$.

Anyway, and since $p>2$ , for simplicity of the proof we'll prove that in fact $2\left(1+\frac{1}{2}+...+\frac{1}{p-1}\right)\equiv 0\!\!\!\pmod {p^2}$

Now, we can form pairs $\frac{1}{i}\,,\,\frac{1}{p-i}\,,\,1\leq i\leq p-1\,\,and\,\,\frac{1}{i}+\frac{1}{p-i}=\frac{p}{i(p-i)}$ , so $2\left(1+\frac{1}{2}+...+\frac{1}{p-1}\right)=2\left(\frac{p}{1(p-1)}+\frac{p}{2(p-2)}+...+\frac{p}{\left(\frac{p-1}{2}\right)\left(\frac{p+1}{2}\right)}\right)=$

$=p\sum\limits_{i=1}^{p-1}\frac{1}{i(p-i)}=p\sum\limits_{i=1}^{p-1}\frac{1}{-i^2}\,\;since\,\;p-i\equiv -i\!\!\!\pmod p$ , so we have to prove this last sum is divisible by p (the minus sign never minds: we can take it out of the sum).

But $\sum\limits_{i=1}^{p-1}\frac{1}{i^2}=\sum\limits_{i=1}^{p-1}i^2\!\!\!\pmod p$ as the only elements in $\mathbb{Z}_p^{*}$ which are inverses to themselves are 1 and -1 and without these two we're left with an even number of

elements, from 2 to p-2 (and here enters the assumption $p\geq 5!$), where we can pair each one with its inverse. Now, the well-known formula

$\sum\limits_{i=1}^{p-1}i^2=\frac{(p+1)(2p+1)}{6}p\,$ means the sum equals zero modulo p since, again, p is prime greater than 5. Q.E.D.

Tonio