Problems & Puzzles: Puzzles

 Puzzle 318. 31 = sigma(16) = sigma(25) Faride Firoozbakht sent the following puzzle: 31 = sigma(16) = sigma(25) Q1. what is the next prime that equals to sigma(n) for two numbers n. Q2. Does there exist a prime p, such that p equals to sigma(n) for three numbers n.

Contributions came form Adam Stinchcombe & J. K. Andersen

Since sigma is multiplicative on relatively prime numbers, sigma(n) can be prime only when n is a power of a prime.  Furthermore, sigma(p^k)=1+p+p^2+..+p^k=[p^(k+1)-1]/[p-1] so if (k+1) factors, say k+1=ab, then sigma(p^k)=([p^a]^b-1)/(p-1)=([p^a]^b-1)/(p^a-1)*(p^a-1)/(p-1) is a composite integer.  Consequently, k+1 is prime.

In other words, we are looking for four primes p,q,r,s with (p^q-1)/(p-1)=(r^s-1)/(r-1) with the shared value also a prime.  The example given corresponds to (2^5-1)/(2-1)=(5^3-1)/(5-1).

I searched through q=3 with the following technique: write M=(r^s-1)/(r-1), so our equation reads (p^3-1)/(p-1)=p^2+p+1=M.  Then an integral value for p requires the discriminant=4M-3 be a perfect square.  I checked for prime s values between 5 and 41 and primes r up to (so far) a little over 13 million, cacluated M, calculated 4M-3, then tested whether this was a quadratic residue mod all primes between 11 and 193.  So far I have found no example for the q=3 case within these constraints.

There was one almost example: if you think of sigma([-91]^2)=[-91]^2+[-91]+1=8191=2^13-1 and 8191 is prime.

Oh well, probably time to think about higher q values........

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Andersen wrote:

If a and b are relatively prime then sigma(a*b) = sigma(a)*sigma(b).
This means n must be a prime power: n = p^a.
sigma(p^a) = p^a + p^(a-1) + ... + p + 1 = (p^(a+1)-1)/(p-1).

sigma(p^a) must be odd to be prime. If p is odd then a must be even.
A computer search shows sigma(2^4) = sigma(5^2) is the only small solution.
I believe it is the only solution at all but I have not made a formal proof.

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