Why $ac=b^2$ forces $a,c$ to be squares if $a,c$ are coprime?

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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.

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

score 10 · Answer 1 · answered Feb 27 '12 at 15:38

10

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 $c$.

answered Feb 27 '12 at 15:38

GeoffDS

11,270

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? – GeoffDS May 01 '12 at 20:04

score 8 · Accepted Answer · edited Apr 13 '17 at 12:21

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.

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

score 2 · Answer 3 · answered Feb 27 '12 at 15:40

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.

Why $ac=b^2$ forces $a,c$ to be squares if $a,c$ are coprime?

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