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
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
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.
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.