The fundamental theorem of arithmetic in number theory states that every positive integer number greater than 1, \(x \in \mathbb{Z}\) where \(x > 1\), is a prime number or it is comprised of the unique product of prime numbers up to the order of the factors.

Example:

\(1250 = 2^{1} \times 5^{4} = 2\times 5\times 5\times 5\times 5\\\)

\(1200 = 3^{1}\times 5^{2}\times 2^{4} = 3\times 5\times 5\times 2\times 2\times 2\times 2\\\)

In other words:

1. 1250 and 1200 can be represented as a product of primes
2. there will always be a specific amount of primes in each product

Proof:

Using Euclid's lemma and B`ezout's identity. Let \(p\) be a prime. Assume \(p\) divides the product of two integers \(a\) and \(b\) written as \((p|ab)\), and its negation is \(\neg (p|ab)\).

Then either \(p\) divides \(a\) \((p|a)\) or \(p\) divides \(b\) \((p|b)\) or both. Some equivalent statements:

• if \(\neg (p|a)\) and \(\neg (p|b)\), then \(\neg (p|ab)\).
• if \(\neg (p|a)\) and \((p|ab)\), then \((p|b)\).

Generalization of Euclid's lemma from prime numbers to any integers:

• if \((n|ab)\), and \(n\) shares no common divisors other than 1 with \(a\), then \((n|b)\).

If \(n\) is prime, either:

• \((n|a)\) or
• \(n\) shares no common divisors other than 1 (i.e. relatively prime) with a, therefore, \(\neg (n|a)\) so \((n|b)\).

Thus, this is a generalization.

B`ezout's identity states that if \(a\) and \(b\) share no common divisor other than 1 (i.e. are relatively prime integers) then there exist integers \(\delta\) and \(\gamma\) such that:

\(\delta a + \gamma b = 1\).

Let a and b share no common divisor other than 1, and assume that \((n|ab)\). According to B`ezout's identity there exist \(\delta\) and \(\gamma\) which provide:

\(\delta n + \gamma a = 1\).

Multiplying both sides by \(b\):

\(\delta nb + \gamma ab = b\).

\(\delta nb\) is divisible by \(n\) and \(\gamma ab\) is divisible by \(ab\) which in turn is also divisible by \(n\) as we state above in our assumption.

Therefore, since we have shown that both terms in the addition of \(\delta nb + \gamma ab\) are divisible by \(n\) then it must be true that their sum \(b\), is also divisible by \(n\).

This is also a generalization to Euclid's lemma mentioned above.