Question #2579

Prove that if two distinct primes p and q divide n^3, then n^3 is a multiple of (pq)^3.

Expert's answer

The number n can be uniquely represented as a product of primes up to their permutation.

Let

n= p1 * p2 * … * pk,

be such a presentation. Some of multiples may repeat.

Then n^{3} has the following presentation

n^{3}= p1^{3} * p2^{3} * … * pk^{3}=

=p1*p1*p1*p2*p2*p2* … *

*pk*pk*pk

Hence if p divides n^{3}, then it divides n, and therefore p^{3} divides n^{3.}

Similarly for q.

Since p and q are distinct numbers, it follows that

(pq)^{3} = p^{3} q^{3}

also divides n^{3}

Let

n= p1 * p2 * … * pk,

be such a presentation. Some of multiples may repeat.

Then n

n

=p1*p1*p1*p2*p2*p2* … *

*pk*pk*pk

Hence if p divides n

Similarly for q.

Since p and q are distinct numbers, it follows that

(pq)

also divides n

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