Prove that if $\displaystyle (a,b)=1$ $\displaystyle ab=c^2$ then there exists integers m and n such that $\displaystyle a=m^2 b=n^2$ I know you have to use prime power decomp to show this but I am stuck. Trying to prep for a final. Thanks

Printable View

- Aug 4th 2009, 01:53 PMdiddledabble[SOLVED] Prime Power Decomposition
Prove that if $\displaystyle (a,b)=1$ $\displaystyle ab=c^2$ then there exists integers m and n such that $\displaystyle a=m^2 b=n^2$ I know you have to use prime power decomp to show this but I am stuck. Trying to prep for a final. Thanks

- Aug 4th 2009, 02:14 PMGamma
well you have pretty much already done it.

Write out the prime factorization of c, then square it and you have a bunch of primes to even powers. Now since $\displaystyle c^2=ab$ and factorization is unique, the factors have to be distributed among a and b. a and b cannot share any of the prime factors or else they would not be relatively prime since that prime would divide both of them. Thus each a and b is a product of even powered prime numbers, making them perfect squares. - Aug 5th 2009, 08:32 AMdiddledabble
I don't follow your logic with factoring c.

- Aug 5th 2009, 08:58 AMGamma
The integers are a Unique Factorization Domain (in fact they are a Euclidean Domain). Every nonzero, and non-unit(things that are invertible, $\displaystyle \pm1$ in the integers) can be written as a product of irreducible elements (primes in this case since in PIDs and UFDs they are the same) unique up to multiplication by a unit (recall, that is$\displaystyle \pm 1$).

This means every integer bigger than 1 has a factorization into primes which is unique up to multiplying it by like 6 negative ones, or 14 negative ones and 7 ones. Get it?

So let $\displaystyle c\in \mathbb{Z}$. If $\displaystyle c=\pm 1$, you are done $\displaystyle 1=(\pm 1)^2=1^21^2=1$ and gcd(1,1)=1.

0 is also clear.

So now we know c is not a unit or zero, so write out its prime factorization.

$\displaystyle c=p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n}$.

$\displaystyle c^2=p_1^{2e_1}p_2^{2e_2}\cdots p_n^{2e_n}$.

Now if this is equal to ab two integers, they must have the same prime factorization. so suppose WLOG a is not a square, then one of those prime exponents must be odd, which means b must have an odd powered exponent. But that means that prime call it $\displaystyle p_i$ divides both a and b, so $\displaystyle 1=gcd(a,b)\geq p_i>1$a contradiction, so both a nd b must be squares.