At the Thirteen Czech-Slovak International Number Theory Conference in Ostravice in 1997 and at JA in Limoges in 1997 A. Schinzel proposed the following problem.
PROBLEM 1. Disprove the following statement.
There exists such a prime number p0, that for all prime numbers p > p0 and all n∈ℕ the following condition holds
2n ≡ 3 mod p ⇔ 3n ≡ 2 mod p.
We can reformulate Problem 1 in the following way.
Prove that for every prime number p0 there is a prime number p > p0 and there is an n∈ℕ such that either
2n ≡ 3 mod p and 3n ≢ 2 mod p
or
2n ≢ 3 mod p and 3n ≡ 2 mod p.
We solve Problem 1 by proving the following theorem.
THEOREM 1.
(a) For every prime number p0 there is a prime number p > p0 and there is an n∈ℕ such that
2n ≡ 3 mod p and 3n ≢ 2 mod p.
(b) For every prime number p0 there is a prime number p > p0 and there is an n∈ℕ such that
3n ≡ 2 mod p and 2n ≢ 3 mod p.
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Vol. 12 (1998)
Published: 1998-09-30
10.1515/amsil
English