Prove that if then there exists integers m and n such that I know you have to use prime power decomp to show this but I am stuck. Trying to prep for a final. Thanks

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- August 4th 2009, 02:53 PMdiddledabble[SOLVED] Prime Power Decomposition
Prove that if then there exists integers m and n such that I know you have to use prime power decomp to show this but I am stuck. Trying to prep for a final. Thanks

- August 4th 2009, 03:14 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 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. - August 5th 2009, 09:32 AMdiddledabble
I don't follow your logic with factoring c.

- August 5th 2009, 09:58 AMGamma
The integers are a Unique Factorization Domain (in fact they are a Euclidean Domain). Every nonzero, and non-unit(things that are invertible, 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 ).

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 . If , you are done 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.

.

.

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 divides both a and b, so a contradiction, so both a nd b must be squares.