Algebra proof involving perfect squares

[johnlin22]

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 available about the GCD of two numbers we have is that it can be written as a linear combination. so ax + by = 1 for some integers x and y.

3) however this problem is in the section about prime numbers and prime factorizations so i started thinking about perfect squares prime factorization

4 = 2^2
9 = 3^2
16 = 2^5 = 2^2(2^2)
25 = 5^2
36 = 2^2(3^2)
49 = 7^2
64 = 2^6 = 2^2(2^2)(2^2)
81 = 3^4 = 3^2(3^2)
and so on. so it seems this this should help me somehow. But i am not sure where to go with this or how i am going to use that fact that (a,b) = 1, so they are relatively prime.

Well wait if they are relatively prime, they couldn't be 2 perfect squares like 36 and 81 or else their GCD would not be 1 it would 3^2 or 9.
But a and b could be numbers like 25 and 16.

Anyone can help me finish please? A lot of these ideas came up as i type this.

 

[betel]

Factorization seems a good way to go. I would try to find out how often a prime factor of c has to appear in the product of ab.
Then the GCD should tell you were this factor can come from in ab.

What's wrong with 25 and 16? They are a solution for c = 20.

 

[johnlin22]

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 some trouble putting this all together though.

Is there anymore help i could get?

 

[betel]

Do it the other way round to get a feeling what you have to do.
E.g. try to find some a,b for c=15.

 

[unusualname]

write ab as a product of distinct prime powers, they must all be even powers if ab = c^2.

Now if gcd(a,b)=1 then a and b are coprime, so can't contain any prime factors in common...

 

[johnlin22]

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?

so
a = (p1^2k1)(p2^2k2)...(pr^2kr)
b = (pr+1^2kr+1)(pr+2^2k+2)...(pn^2kn)

aha so then a and b are perfect squares. this is right?

does this work?