Let me start with C.
c = p1^k1(p2^k2)...pn^kn.
c^2 = p1^2k1(p2^2k2)...pn^2kn
and we know (a,b) = 1 are relatively prime so they do not share any of the same p sub i right?
But i can definitely say that the prime factorization of c^2 better be the same the prime factorization of ab right...
I am still a bit confused. So just as an example if a were 25 and b were 16 then that's fine because (a,b) = 1.
so the c = 20 we have (20)^2 = 25(16) and so we have 400 = 400.
20 = 2^2(5).
so now i am starting to see something. c has the same prime factorization has a*b. I am having...
The problem states, if c^2 = ab and (a,b) = 1, prove that a and b are perfect squares.
( the notation (a,b) means the GCD of a and b)
So i have a lot of thoughts on this problem but i am getting stuck.
1) if c^2 = ab then c(c) = ab which says c divides ab.
2) the only real theorem...