Let k be in Z and k>0. Prove that the equation sum of positive divisors function(n)=k has at most finitely many solutions.
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Well, for any integer n, the sum of the positive divisors of n is at least n+1.
Originally Posted by Tinyboss Well, for any integer n, the sum of the positive divisors of n is at least n+1. The sum of positive divisors(n) $\displaystyle \leq 1+2+\cdot\cdot\cdot +n = \frac{n(n+1)}{2} $. NEVERMIND, IGNORE THIS POST I READ THE QUESTION WRONG
No one knows this one? I am still having trouble with it.
Originally Posted by meshel88 No one knows this one? I am still having trouble with it. Tinyboss answered this pretty well.
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