# Prove that $1$ has only one divisor

I'm looking at Euclid's Theorem (the infinitude of primes).

The standard proof assumes there are finitely many primes (and proceeds to contradiction). It involves $P :=$ the product of all the primes, and $Q := P+1$.

Since every prime divides $P$, no prime can divide $Q$; since if something divides two numbers it must also divide their difference, and the difference between $P$ and $Q$ is $1$, and nothing divides $1$ except $1$.

But... why does nothing divide $1$ except $1$? Given the definition of divisibility

$$a\mid b \iff \exists x : ax = b$$

how can you prove that $1$ has only one divisor?

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Euclid's actual proof is not a proof by contradiction (this is discussed to no end on the site, e.g. at math.stackexchange.com/a/632129/264) – Zev Chonoles Feb 18 '14 at 6:23
Also, the number $-1$ is also a divisor of $1$, since $(-1)(-1)=1$. – Zev Chonoles Feb 18 '14 at 6:24

If $1=de$ with $d,e$ positive integers, you can't have them both bigger than 1 (since their product would be bigger than 1), nor can either one be less than 1 (since there are no integers between 0 and 1), so one of them must be 1 (and so the other must be 1).
$a\gt1$ implies $ab\gt b$; if also $b\gt1$ then $ab\gt1$. In the end, though, the question of how to prove things comes down to the question of what set of axioms you are starting with. – Gerry Myerson Feb 19 '14 at 1:47