# 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 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.

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every number can be represented as a product of powers of primes – dato datuashvili Feb 27 '12 at 15:40

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.

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Thank you very much for your deep explanation. I now have a better understanding. – T W Lai Feb 27 '12 at 17:35

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 $c$.

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Thank you very much for your clear and concise explanation. It is very helpful. – T W Lai Feb 27 '12 at 17:22
Would the downvoter, 2 months after the answer, like to explain? – Graphth May 1 '12 at 20:04

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.

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