Show that if $\gcd(p, a) = \gcd(p, b) = 1 \implies \gcd(p, ab) = 1$, then $p$ is prime.

Question

Show that if $$\gcd(p, a) = \gcd(p, b) = 1\implies \gcd(p, ab) = 1$$ for all $a$ and $b$, then the only divisors of $p$ are $\pm 1$ and $\pm p$, i.e. $p$ is prime (when $p \neq 0, \, \pm 1$).

Answer 1

The statement $$\forall a\forall b(\gcd(n,a)=1\wedge\gcd(n,b)=1\Rightarrow\gcd(n,ab)=1)\Rightarrow\text{$n$ prime}$$ is false for $\gcd(6,a)=\gcd(6,b)=1$ implies $\gcd(6,ab)=1$ for if $q\in\mathbb N$ is a prime and $q\mid\gcd(6,ab)$ then $q=2$ or $q=3$ and $q\mid a$ or $q\mid b$ and this is impossible since $\gcd(2,a)=\gcd(3,a)=\gcd(2,b)=\gcd(3,b)=1$.

The correct statement is:

If for all $a $ and $b $, $p\nmid a $ and $p\nmid b $ implies $p\nmid > ab $, then $p $ is prime

that's $$\forall a\forall b(n\nmid a\wedge n\nmid b\Rightarrow n\nmid ab)\Rightarrow\text{$n$ prime}$$

Answer 2

There is something wrong in the statement, as the implication

$\gcd(c, a) = \gcd(c, b) = 1\implies \gcd(c, ab) = 1$

holds true generally for all $a, b, c$, so it is an empty condition on $c$.

This is because, by Bézout's lemma, if $\gcd(c, a) = \gcd(c, b) = 1$ then there are integers $x, y, z, t$ such that $$ c x + a y = 1 = c z + b t, $$ so that $$ 1 = c x + a y (c z + b t) = c (x + a y z) + a b (y t), $$ which implies $\gcd(c, a b) = 1$.

Answer 3

HINT: consider the fundamental theorem of arithmetics. Given that you know $p, a $'s factorisation, how does one calculate $\gcd(p,a) $? If the $\gcd $ is 1, what does that tell about $p $ and $a $'s factorisation relating them with eachother? The same for $p, b $. What is the factorisation of $ab $? Given what we know about $p $ relating to $a $ and $b $, can $ab $ have a divisor that divides $p $?

That is, write $p, a $ and $b $ as their unique product of primes raised to some exponents. If one knows the factorisation of $c,d $ then finding $\gcd(c,d) $ is a trivial task. Can you tell why? If you can, your statement should become evident as soon as you take into account what the left side of the implication tells about the factorisations of $p, a $ and $b $.

EDIT: made the answer more concrete

First note that, if $a,b,c \in \Bbb Z $ are such that $\gcd(c,a)=\gcd(c,b)=1$, then automatically we have $\gcd (c,ab) = 1$. Regardless of $c $ being prime or not. You can either prove it with a similar method as what I am about to do or refer to @Andreas Caranti's answer.

Therefore, the right side of the implication is irrelevant since it will always be true.

You are now left with

If $\gcd(p, a) = 1$ for all $a $ not multiple of $p, p \not= 1$, then $p $ is prime.

But that must be true. Recall the fundamental theorem of arithmetics: each integer has a unique factorisation as the product of primes, up to a permutation of the factors.

Also, suppose that $a = a_1^{e_1}\cdots a_i^{e_i}, b = b_1^{f_1}\cdots b_j^{f_j} $ are the factorisations of two integers $a $ and $b $. If $ \gcd(a,b) = d > 1$, then $d $ also has some factorisation, and what is more, its factors appear in the factorisations of $a $ and $b $! For example, suppose $d_k^{g_k} $ is in the factorisation of $d $. Then, there is some $k_1$ such that $a_{k_1} = d_k, e_{k_1} \geq g_k $ and some $k_2$ such that $b_{k_2} = d_k, f_{k_2} \geq g_k $. This allows you to calculate GCDs very easily provided you know the prime factorisations of the numbers (can you tell how?).

You want to prove $\forall a \in \Bbb Z \setminus \{kp: k \in \Bbb Z\}, \gcd(p, a) = 1 \implies p$ prime.

Let us prove that by showing that if $p$ is not prime, then there exists such an $a \not=1$ such that $\gcd(p,a) > 1$. Say $p $ is not prime and let $p = p_1^{q_1}\cdots p_n^{q_n} $ be its factorisation. Pick any $p_i $ found there. $\gcd(p, p_i) = p_i > 1$ because $p_i$ is prime.

Answer 4

If $a,b < p$ then $p$ is prime if all the numbers less than $p$ are coprime with $p$: $$\sum_{i=1}^{p-1}\gcd(i,p) = p-1$$ This is the same as Euler totient function that states that $p$ is prime $\iff \varphi(p)=p-1$

Note that if $a,b > p$ a counter-example would be $a=5, b=7, p=6$ yielding $\gcd(6,5)=1 \gcd(6,7) = 1 \gcd(6,35)=1$