proof explanation - If $p_1,...,p_n$ are positive prime numbers then $sqrt{p_1cdots p_n} notin Bbb{Q}$

We want to proove that:

If $p_1,...,p_n$ are positive, distinct, prime numbers then
$\sqrt{p_1\cdots p_n} \notin \Bbb{Q}$.

Let's assume that $\sqrt{p_1\cdots p_n} \in \Bbb{Q}$. Then, $\exists (a,b)\in \mathbb{Z^*\times Z^*}:\sqrt{p_1\cdots p_n}=\frac{a}{b}$ with $\gcd (a,b)=1.$ So, $a^2=p_1\cdots p_n \cdot b^2. $ But how do we continue? Is this technique right or should we follow something different?

PS: This is a part this proof, and I would like to discuss it.

Thank you.

Answer

The classic proof that $\sqrt 2 \notin \Bbb Q$ readily extends to this case, to wit:

if

$\sqrt{p_1p_2 . . . p_n} = \dfrac{a}{b} \tag 1$

with $a, b \in \Bbb N$, $\gcd(a, b) = 1$, then

$p_1p_2 \ldots p_n b^2 = a^2, \tag 2$

whence $p_1 \mid a^2$; thus $p_1 \mid a$ and so $a = p_1c$; thus

$a^2 = p_1^2c^2 = p_1p_2 \ldots p_n b^2, \tag 3$

whence

$p_1c^2 = p_2p_3 \ldots p_n b^2; \tag 4$

since the $p_i$ are distinct, we must have $p_1 \mid b^2$, whence $p_1 \mid b$, contradicting our assumption that $\gcd(a, b) = 1$.

I think the ancient roots or this proof are worthy or respect, and I also note the method readily extends to show many similar propositions hold.

I presented this answer because it seems to me that the linked proof is pretty complex for this particular problem, although it is certainly engaging in and of itself, and leads in entaging directions.