Problems & Puzzles: Conjectures
Conjecture 97. Another conjecture related to Mersenne primes
Sebastian Martín Ruiz sent the following conjecture:
Let p a prime number.
If p+1=2^(k1)(q2)^(k2)....(qn)^(kn) is the prime number decomposition
of p+1 where qi are odd primes i=2,...,n we have
(qi)^(ki) divides p+2^Phi(p+1)
that is to say p+2^Phi(p+1) is a composite number except at most
in the case that p is a Mersenne prime.
I have not found more primes.
Q1 Prove the conjecture or find a counterexample.
Q2 Find more prime values of p+2^Phi(p+1) searching for p Mersenne primes.
During the week from March 25-31, 2023 contributions came from Emmanuel Vantieghem and Oscar Volpatti
Oscar wrote:Conjecture 97 is true.
Function F(n) = n+2^Phi(n+1) is well defined for every non-negative integer n.
Consider sequence M(k) = 2^k-1 (OEIS A000225, sometimes called Mersenne integers).
THEOREM: if n doesn't belong to sequence M(k), then F(n) is composite.
Let's distinguish two cases, according to the parity of n.
CASE 1: n is odd.
Then n+1 is even; let 2^e be the largest power of 2 dividing n+1, so that n+1 = d*2^e, with d odd.
Cofactor d must be greater than 1, otherwise n = 2^e-1 = M(e), belonging to sequence M(k).
Define constant b = 2^Phi(2^e) = 2^(2^(e-1)), clearly coprime with d.
Totient function is multiplicative, so we can rewrite term 2^Phi(n+1) as follows:
2^Phi(n+1) = 2^(Phi(d)*Phi(2^e)) = (2^Phi(2^e))^Phi(d) = b^Phi(d).
But b^Phi(d) == 1 mod d by Euler's totient theorem, so F(n) == n+1 == 0 mod d.
Hence, d is a proper factor of F(n).
CASE 2: n is even.
Then n is not smaller than 2, otherwise n = 0 = 2^0-1 = M(0), belonging to sequence M(k).Inequality Phi(n+1) >= 2 holds for every integer n >= 2, because numbers 1 and n are distinct and both coprime with n+1.
2^Phi(n+1) is even too, and not smaller than 2^2 = 4;
F(n) is even too, and not smaller than 2+4 = 6.
Hence, 2 is a proper factor of F(n).
Let's restrict our attention to arguments n belonging to sequence M(k).
For compactness, define function G(k) = F(M(k)) = 2^k-1+2^Phi(2^k).
G(k) is prime for exponents 0 <= k <= 4:
G(0) = 2,
G(1) = 3,
G(2) = 7,
G(3) = 23,
G(4) = 271.
For larger k, trial factoring up to 2^32 often suffices to detect proper factors of G(k).
The first few exceptions are exponents 15, 18, 23, 51, 60, 73, 78, 84, 89...Some small prime factors appear frequently, with simple divisibility rules:7 divides G(k) iff k == 2 mod 6;
11 divides G(k) iff k == 7 or 16 mod 20;
13 divides G(k) iff k == 7 mod 12;
19 divides G(k) iff k == 11 or 13 mod 18;
23 divides G(k) iff k == 3, 17, 20, 59, or 76 mod 110;
29 divides G(k) iff k == 34, 50 or 69 mod 84;
31 divides G(k) iff k == 19 mod 20;
and so on.About exceptions, PRP test proved both G(15) and G(18) as composites.Function G(k) is strictly increasing: G(23) has 1262612 digits, while G(51) has 3.389*10^14 digits!Therefore, k=23 seems to be last exponent for which a PRP test is currently feasible, but it would still take ages of work on my laptop.For larger exponents, the only chance is to prove G(k) as composite by finding some small factor.
As an example, trial factoring up to 10^11 allows to discard some more exponents:
4599739031 divides G(60),
55084086769 divides G(89)...
Let's further restrict our attention to prime arguments p belonging to sequence M(k); any such prime is by definition a Mersenne prime, so I checked the exponents of all 51 known Mersenne primes. After prime values G(2) and G(3), trial factoring up to 2^32 detected proper factors of G(k) for 47 exponents, the only exceptions being k=89 (solved) and k=13466917 (no factors below 10^11).
See attached files C97factorsSMALL.txt and C97factorsPRIMES.txt for my factorization results.
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