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

11,410

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

Bill Dubuque · Accepted Answer · 2024-06-18T19:15:16.710

It's easy by prime factorization. More generally by gcds we explicitly show $\rm\:a,c\:$ are squares:

Lemma $\rm\,\ \ \color{#0a0}{(a,b,c) = 1},\,\ \color{#c00}{b^2 = ac}\ \Rightarrow\ a = (a,b)^2,\ c = (c,b)^2\ $ for $\rm\:a,b,c\in \mathbb N$

Proof $\rm\ \ (a,b)^2 = (a^2,ab,\color{#c00}{b^2}) = (a^2,ab,\color{#c00}{ac}) = a\color{#0a0}{(a,b,c)} = a.\,$ Same for $\rm\,(c,b).\,$ $\ \bf\small QED$

Yours is the special case $\rm\:(a,c) = 1\ (\Rightarrow\ (a,b,c) = 1)$. Note that the above proof uses only basic gcd arithmetic, i.e. universal gcd laws (associative, commutative, distributive) therefore it generalizes to any gcd domain/monoid, e.g. any UFD.

Generally $\rm\, ac = bd\Rightarrow (a,b)(a,d) = (aa,ab,bd,ad) = a\, \color{#0a0}{(a,b,c,d)} = a\,$ if $\rm\,\color{#0a0}{(a,b,c,d) = 1}.\:$ For more on this and related topics such as Euler's Four Number Theorem (Vierzahlensatz), Riesz interpolation, or Schreier refinement see here and here .

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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