Product of Two Distinct Primes has 4 Positive Divisors

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Let $n \in \Z_{>0}$ be a positive integer which is the product of $2$ distinct primes.

Then $n$ has exactly $4$ positive divisors.


Let $n = p \times q$ where $p$ and $q$ are primes.

We have by definition of divisor:

\(\ds 1\) \(\divides\) \(\ds n\) One Divides all Integers
\(\ds p\) \(\divides\) \(\ds n\) Definition of Divisor of Integer
\(\ds q\) \(\divides\) \(\ds n\) Definition of Divisor of Integer
\(\ds p \times q\) \(\divides\) \(\ds n\) Integer Divides Itself

where $\divides$ denotes divisibility.

Suppose $a \divides n$ such that $1 \le a < n$.

Suppose $a \ne p$.

By definition of prime number:

$a \perp p$

where $\perp$ denotes coprimality.

From Euclid's Lemma:

$a \divides q$

and so by definition of prime number:

$a = q$


$a = 1$

Similarly, suppose $a \ne q$.

By the same reasoning:

$a = p$


$a = 1$

Thus the only positive divisors are as above.