Why $ac=b^2$ forces $a,c$ to be squares if $a,c$ are coprime? I was browsing this post:
Prove that $a^2  + b^2 + c^2 $ is not a prime number
One of the answer has the following statement:
"If the numbers $a$ and $c$ are coprime, then the equation $ac=b^2$ together with unique factorization forces both $a$ and $c$ to be squares."
I know this should be a trivial question. However, I am not sure how to show the above statement is correct? Any help will be greatly appreciated.
 A: Look at the prime factorizations.  In $b^2$, all primes occur as even powers.  Therefore, in $ac$, they must occur as even powers.  In $ac$, since they are coprime, this is the product of the primes in the factorization of $a$ times the product of the primes in the factorization in $c$, with no overlap, no primes in common.  Therefore, the primes in the factorization of $a$ must have been even powers to begin with, and similarly with $c$.
A: In fact one can explicitly show that $\rm\:a,c\:$ are squares by taking gcds. Namely 
Lemma $\rm\ \ (a,b,c) = 1,\ ac = b^2\ \Rightarrow\ a = (a,b)^2,\ c = (c,b)^2\ $ for $\rm\:a,b,c\in \mathbb N$
Proof $\rm\ \ (a,b)^2 = (a^2,ab,b^2) = (a^2,ab,ac) = a(a,b,c) = a\ \ $ QED
Yours is the special case $\rm\:(a,c) = 1\ (\Rightarrow\ (a,b,c) = 1)$. Note that the above proof uses only universal gcd laws (associative, commutative, distributive) so it generalizes to any gcd domain/monoid.
Generally $\rm\: ac = bd\: \Rightarrow\: (a,b)(a,d) = (aa,ab,bd,ad) = a\: (a,b,c,d) = a\:$ if $\rm\:(a,b,c,d) = 1.\:$  For more on this and closely related topics such as Euler's four number theorem (Vierzahlensatz), Riesz interpolation, or Schreier refinement see this post and this post.
A: Well if you factorise both $a$ and $c$ into primes then the fact that $a,c$ are coprime tells us that there is no common prime in the factorisations.
This means that when you do the product, to get a square you must have an even power of each prime occuring...hence $a,c$ are perfect squares.
