# Show if a product of coprime numbers is a perfect square, so are the numbers - without FTA

I want to prove: $$\text{If }\gcd(a,b)=1\text{ and }ab=n^2,\text{ then }a,b\text{ are also perfect squares.}$$ Assume everyone is a positive integer, etc. Unless I'm deluding myself, this is pretty easy to show using unique prime factorization.

But I want to do it without using primes or the (usual statement of the) FTA. That is, using coprime is fine, using the so-called Bezout identity (XGCD algorithm), etc. is fine. Is this even possible without essentially defining at least irreducibles, if not primes and prime factorization, along the way?

(See here for a more vague question I asked a while ago on this.)

• See my answer here which uses only properties of gcds. Note the in $\Bbb Z$ (or any domain where nonzero nonunits factor into irreducibles), the existence of gcds is equivalent to uniqueness of factorization into irreducibles, since it easily implies that irreducibles are prime. – Math Gems Jan 24 '13 at 20:10
• A very related question: Let $R$ be a Bezout domain. If $a,b$ are coprime and $ab$ is a square, does it follow that $a,b$ are squares ? – Amr Jan 24 '13 at 20:21
• @Amr Yes, Bezout domains are gcd domains, so the above-linked proof works. – Math Gems Jan 24 '13 at 20:24
• @MathGems Yes I know. I was just telling the OP that this may be a better way to ask the question – Amr Jan 24 '13 at 20:32
• @MathGems - nice, yes you are right about this. Maybe I should just be asking about Bezout domains instead - haven't thought about the tower of domains in years, thanks. I up-voted MathGems' earlier answer, best I could do here. – kcrisman Jan 24 '13 at 20:58

According to this meta question and this meta answer, I will post @MathGems answer here as community wiki just so it's easier to find. (If they return they can post and I'll delete this.)Directly quoted:

Alternatively, and more generally, one can use gcds to explicitly show $\rm\:a,b\:$ are squares. Writing $\,\rm(m,n,\ldots)\:$ for $\rm\: gcd(m,n,\ldots)$ we have:

Theorem $\rm\ \ \color{#C00}{c^2 = ab}\, \Rightarrow\ a = (a,c)^2,\ b = (b,c)^2\:$ if $\rm\ \color{#0A0}{(a,b,c)} = 1\$ and $\rm\:a,b,c\in \mathbb N$

Proof $\rm\ \ \ \ (a,c)^2 =\: (a^2,\color{#C00}{c^2},ac)\: =\: (a^2,\color{#C00}{ab},ac)\: =\: a\,\color{#0A0}{(a,b,c)} = a.\:$ Similarly $\rm \,(b,c)^2 = b.\ \$ QED

Yours is the special case $\rm\:(a,b) = 1\ (\Rightarrow\ (a,b,c) = 1)$. The above proof uses only universal gcd laws (associative, commutative, distributive), so it generalizes to any gcd domain/monoid (where, generally, prime factorizations need not exist, hence the above proof is more general).

• It is clear that if $x$ divides $a$ and $c$ then $x^2$ divides each of $a^2,c^2, ac$. So $(a,c)^2\leq (a^2,c^2,ac).$ But how do you show the reverse inequality? And how do you show the second inequality in the proof? – DanielWainfleet Sep 21 '16 at 2:47
• I think this is general property of gcd "arithmetic" (an exercise in some texts?); recall that I'm just copying another answer here for reference. – kcrisman Sep 21 '16 at 12:58