My question reads: If c^2=ab and (a,b)=1, prove that a and b are perfect squares.
I began my proof by $a=p_1 p_2 \cdots p_n$ and $b=q_1 q_2\cdots q_m$. Then I gave $c$ its own decomposition as well and said $c=s_1 s_2\cdots s_t$.
From there I squares the c and just got the same except now with 2 in the exponent for each s.
I am not too sure how to continue on from there. Would I need to re-index? Also, I am not too sure when to bring in the fact that their gcd is 1.
On
Hint $ $ Either compare the parity of the exponents in prime factorizations, or induct as below.
Induct on $\,k =\,$ number of prime factors of $\,c.\,$ If $\,k=0\,$ then $\,c=1\,$ so $\,a=1=b\,$ are squares. Else $\,k\ge 1,\,$ so some prime $\,p\mid c,\,$ so $\,p^2\mid c^2\! = ab\,$ thus $\,p^2\mid a\,$ or $\,p^2\mid b\,$ by $\,a,b\,$ coprime. Wlog $\,p^2\!\mid b,\,$ so canceling $\,p^2$ we get $\,a(b/p^2) = (c/p)^2.\,$ $\,c/p\,$ has less prime factors than $\,c\,$ hence by induction we deduce that $\, a = j^2,\,\ b/p^2 = k^2,\,$ so $\,b = (pk)^2$.
Remark $\ $ The proof generalizes to higher powers and to any UFD. Alternatively a simple proof using gcd laws give an explicit representation, viz. $\ a = \gcd(a,c)^2,\,\ b = \gcd(b,c)^2.$
Pick a prime $p$ that divides $a$, we must show the maximum power of $p$ that divides $a$ is even. But this number is equal to the maximum power of $p$ that divides $c$ (because $c=ab$ and $p$ does not divide $b$), which is even because $c$ is a square.