Maths Olympiad Prep

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Number theory Difficulty 5.8 AIME, harder Prove it

1. Definition of prime numbers and some basic properties An integer nn greater than 1 has at least two distinct positive divisors: 1 and nn. If nn has no divisor greater than 1 and less than nn, then nn is called a prime number. If nn has a divisor greater than 1 and less than nn, i.e., nn can be expressed in the form of aba \cdot b, then nn is called a composite number.
(1) Any integer greater than 1 must have a prime factor.
(2) There is only one positive integer that is both even and prime, which is 2.
(3) Let pp be a prime number, and nn be any integer, then either pp divides nn, or pp and nn are coprime.
(4) Let pp be a prime number, and a,ba, b be integers. If pabp \mid a b, then at least one of a,ba, b is divisible by pp. Proof: If pp does not divide aa and bb, then pp is coprime with aa and bb, thus pp is coprime with aba b, which is a contradiction!
(5) There are infinitely many prime numbers. Prove this proposition by contradiction, assume there are only finitely many prime numbers, let p1=2,p2p_{1}=2, p_{2} =3,,pk=3, \cdots, p_{k} be all the prime numbers, consider N=p1p2pk+1N=p_{1} p_{2} \cdots p_{k}+1, clearly N>1N>1. Therefore, NN has a prime factor pp. Since p1,p2p_{1}, p_{2}, ,pk\cdots, p_{k} are all the prime numbers, pp must equal some pi(1ik)p_{i}(1 \leqslant i \leqslant k), thus pp divides Np1p2pk=1N-p_{1} p_{2} \cdots p_{k}=1, which is impossible, therefore there are infinitely many prime numbers.

Solution

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.