Problem on a square group Let G be a finite group. 
An element $a \in  G$ is called a square if there exists $x \in G$ such
that $x^ 2 = a$.
Is the following statement(s) is/are true?
(A) If $a,b ∈ G$ are not squares, $ab$ is a square.
(B) Suppose that $G$ is cyclic. Then if $a,b ∈ G$ are not squares, $ab$ is a square
A very weird problem.Neither able to find counter examples nor able to prove it.What to do?
 A: When working with additive notation being a square can be translated to being even (being the sum of an element with itself)
Part $A$:
Take $\mathbb Z_2\times \mathbb Z_2$ with additive notation.
$(1,0)$ is not a square since $(0,0)$ is the only square (adding an element to itself gives $(0,0)$
$(0,1)$ is not a square.
$(1,1)=(1,0)+(0,1)$ is not a square, so it is false.

Part $B$:
Notice that this is true in $\mathbb Z$ (sum of odd numbers is even). Since all infinite cyclics are isomorphic to $\mathbb Z$ we are done.
We now go to $\mathbb Z_n$.
if $n$ is odd then every element is a "square" since if $m$ is even $m$ then $m=\frac{m}{2}+\frac{m}{2}$ (where by $\frac{m}{2}$ I mean actually taking the integer $\frac{m}{2}$ and then considering the residue class). On the other hand if $m$ is odd then $n+m$ is even and so we can do the same thing, taking into account in $\mathbb Z_n$ $m$ and $n+m$ are the same thing.
If $n$ is even then the "squares" are exactly the elements that are even, this is because reducing a number $\bmod$ an even modulus preserves the parity, in this case we just use that odd plus odd is even, and so the sum of two non-squares is a square.
A: I have a partial answer for the proof. part b. If G is cyclic generated by g and a is not a square, then we can find an integer k such that $a=g^{2k+1}$, and similarly we write $b=g^{2q+1}$, $ab=g^{2k+1+2q+1}=g^{2(k+q+1)}=(g^{k+q+1})^{2}$. 
A: Notice that in any abelian group $G$, the set of squares $S$ is a subgroup, since $x^2y^2=(xy)^2$. In fact, it is a normal subgroup (since all subgroups are normal in abelian groups), which means we can consider the quotient group $G/S$ of multiplication of cosets. One can quickly take that:

A cyclic group is generated by one element. Let $x$ be such an element. The quotient $G/S$ must also be generated by one element, call it $y$. Since all squares are in $S$, it follows that $y^2=e$. Thus $G/S$ is either the cyclic group on two elements or the trivial group.

From there, we have that if $x$ and $y$ are non-squares, they are in the same coset of $S$ (since there are at most two cosets, one being $S$ itself). This coset has order two hence squares to $S$ - in particular, meaning that $xy$ is a square.
If we have an abelian group of more than one generator (for instance, in $\mathbb Z\times\mathbb Z$), then the quotient $G/S$ can have more than one generator - meaning non-square $x$ and $y$ might not belong to the same coset and the product of the cosets might not be $S$.
A: Another counter example for part $A$:
Consider $Q_8=\{\pm1,\pm i,\pm j,\pm k\}$.  Here $i$ and $j$ are not squares, because any $x \in Q_8$ satisfies either $x^2=1$ or $x^2=-1$ .  Also $ij=k$ is not a square!
