RE A114705

[Tautócrona]

----- Original Message ----- 
From: "zak seidov" <zakseidov at yahoo.com>

All terms are even?

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Yes, they are:

p= 2^n+3^n is an odd number. It should have an even number of divisors to make sigma even.

The number of divisors of the number p = p1^e1 · p2^e2 · ... · p_k^e_k is D = (e1 + 1)·(e2 
+ 1)·...·(e_k + 1). If any of the e_k is an odd number, then D is even. So for D to be 
odd, all the e_k have to be even and therefore p must be a square number.

But:

Note that 2^4 = = 1 (mod 15), 3^4 = = 6 (mod 15) and 6^k = = 6 (mod 15)

If n = = 1 (mod 4),
p = = 2^(4k+1) + 3^(4k+1) (mod 15)
p = = 2^1 + 6·3^1 = = 5 (mod 15)

If n = = 2 (mod 4),
p = = 2^(4k+2) + 3^(4k+2) (mod 15)
p = = 2^2 + 6·3^2 = = 13 (mod 15)

If n = = 3 (mod 4),
p = = 2^(4k+3) + 3^(4k+3) (mod 15)
p = = 2^3 + 6·3^3 = = 5 (mod 15)

If n = = 0 (mod 4),
p = = 2^(4k) + 3^(4k) (mod 15)
p = = 1 + 6 = = 7 (mod 15)

Therefore the possible residues for 2^n+3^n (mod 15) are 5,7,13 but all of these are 
quadratic non-residues: the quadratic residues mod 15 are 0,1,4,6,9,10. p can never be an 
quadratic number, so D is an even number and sigma(2^n + 3^n) is an even.

Regards. Jose Brox