(In reply to
A start? by Jer)
Jer,
You did the hard part:
1. Clearly, m never divides m^2m+1
2. Note that m^2m+1 =(m(m1))+1. Since (m1) divides m^2m, it never divides m^2m+1 either.
3. But we also have, for any m, m^2m+1 = (m1)^2+(m1)+1:
(m+1) = (m1)+2
Let (m1) = n:
(n^2+n+1)/(n+2) = (n1)+3/(n+2); integral iff n = {1, 1, 3}
4. m^2m+1+2m = m^2+m+1; as you point out, clearly, these two expressions have no common divisors greater than 1.
So the only candidate is n = 1, i.e. m = 2: 2^61 = 63 = 3^2x7; (2^31)(2^21) = 3*7, satisfying the condition.

Posted by broll
on 20150425 23:49:52 