How does one prove when two cycles in the group of permutations satisfy A^2 = B^2 that A = B?
I think the fact the inverses A_*A = e and A_ = A^(l-1) and A = A^(l+1) is useful. l is the length of the cycles A and B, l is odd. Because V^2 is a cycle if and only if the length is odd.
I have a tendency to assume proofs have to be more complicated then they need to be.
We need to assume cycle lengths of $A,B$ are odd. Otherwise $(1,2,3,4)$ and $(1,4,3,2)$ have the same squares. Maybe you want all of $A,B,A^2,B^2$ to be cycles, then the lengths must be odd. [But that wasn't said in question]
Anyway first show if $A^2=B^2$ the $A,B$ have the same length, say $2n-1.$ Then since $A^{2n-1}=1$ [because cycle to length of cycle is 1] you have $A^{2n}=A.$ Similarly $B^{2n}=B.$ But then
$$A=A^{2n}=[A^2]^n=[B^2]^n=B^{2n}=B,$$ where in the middle we used assumption $A^2=B^2.$