Problems & Puzzles: Puzzles

Puzzle 392. σ(x) = x+φ(x) Farideh Firoozbakht asks for soultions other than x=2, to: σ(x) = x+φ(x) On my request Firoozbakht sent the following graph of f(x) vs x, where f(x) = σ(x) - x - φ(x). Questions. Find solutions or show that these are impossible for x>2.

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Contribution came from Rustem Aidagulov

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Rustem wrote:

It is trivial, that x=y^2, were y is odd. Really, if x is even and x>2, then sigma(x)>3x/2, but phi(x)<=x/2 -contradition. If x is odd, then phi(x) is even, therefore x+phi(x) odd. But sigma(x) is odd only for x=y^2 (x is odd). It mean all prime divisors x are divisors of sigma(x). I think, that is easy to prove, that it is impossible.

It is easy to prove, that y (or x) had at least 3 prime divisors. Let a=Product_(p|x) (1-1/p), then x+phi(x)=(1+a)x, but sigma(x)<x/a. Therefore a(a+1)<1 or a<(sqrt(5)-1)/2=0.6180... If minimal prime p>3 then obviosly y had at least 3 prime divisors.

If y=3^np^m, then a=2(p-1)/3p and equation equivalent to (3-3^{-2n})((p-p^{-2m})=3p+(p^2-14p+4)/3p. Therefore p<=13 and 3*13^{-2m}+13*3^{-2n}-3^{-2n}p^{-2m}=(14-p)/3-4/(3p). It is not hard to check, that it hard not solution.

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