<s>[INST] $p$ prime and $m_p$ a proper divisor of $p−1$. Is it $\sigma(m_p)-\nu(m_p)(This is a sharper version of this other question of mine.)
Along the problem I'm facing, I've come to the following lemma (if it is true):
Let $p$ be a prime and $m_p$ a proper divisor of $p-1$. Then the difference between the sum of all the divisors of $m_p$, say $\sigma(m_p)$, and their number, say $\nu(m_p)$, is less than $p-2$.
Examples:

*

*$p=13$; then, for $m_{13}=6$: $1+2+3+6-4=8<11$, and likewise for $m_{13}=4,3,2,1$;

*$p=31$; then for $m_{31}=15$ we have: $1+3+5+15-4=20<29$, or for $m_{31}=10$: $1+2+5+10-4=14<29$, and likewise for $m_{31}=5,3,2,1$;

*$p=37$; then for $m_{37}=18$ we have: $1+2+3+6+9+18-6=33<35$, and likewise for $m_{37}=9,6,3,2,1$;

*$p=101$; then for $m_{101}=50$ we have: $1+2+5+10+25+50-6=87<99$, and likewise for $m_{101}=25,20,10,5,2,1$.

Is it true for every $p$ and $m_p$?
 [/INST] By computer search, $p = 61$ and $m_p = 30$ seems to be the smallest counterexample. We get $\sigma(m_p) = 1 + 2 + 3 + 5 + 6 + 10 + 15 + 30 = 72$ and $\nu(m_p) = 8$, so $\sigma(m_p) - \nu(m_p) = 72 - 8 = 64$.
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