$A/J$ has a square root and $x-y^2\in J$. 
$A$ is a ring and $J$ is an ideal of $A$. Prove that $A/J$ has a square root iff for every $x \in A$ there is some $y \in A$ such that $x-y^2\in J$.

My proof is rough and I think I have an idea on what to do but I can't quite convey it.
If $J$ has a square root for some $y\in A$ then $J+x=(J+y)(J+y)=(J+y)^2$, so $J+x=J^2+Jy+yJ+y^2$, thus $J+x=J+y^2$ which is $(J+x)+(J-y^2)=J+(x-y^2)$ hence $x-y^2\in J$.
If I start with $x-y^2\in J$ then we know $J+(x-y^2)\in J$ and $J+(x-y^2)=(J+x)+(J-y^2)$ so $J+x=J+y^2$ thus J has a square root.
 A: Our OP Kelly R seems to be on the right track, though certain statements which close the arguments have been omitted, the "mathematical phrasing" is at certain points awkward, and there  appear to be some typographical errors. 
I'll first remark on the apparent typos:  for instance, in the paragraph which begins, "If $J$ has a square root for some $y \in A$ etc", apparently, something like, "If $A/J$ has a square root for some $J + x$  . . . " is really what is intended there; I base this observation on the overall context.  Another example is that I'm pretty sure our OP meant to write $J + (x - y^2) \subset J$ and not $J + (x - y^2) \in J$ in the paragraph which begins
"If I start with $x - y^2 \in J$ . . . "
My second comment addresses the lack of "mathematical closure" at certain points; an example may be seen in the same paragraph referenced above; the reasoning is rightly taken to the point "$J + x = J + y^2$", from which it is correctly if somewhat awkwardly inferred that "$(J + x) + (J - y^2) = J + (x - y^2)$"; missing, however, is the final rejoinder to the effect that $J + (x -y^2) = J$ which allows the essential conclusion $x - y^2 \in J$; this of course follows from $J + x = J + y^2$ since this implies $J + (x - y^2) = J$, equivalent to $x - y^2 \in J$.
The only other thing I would add is that some of the expressions used seem a little cumbersome, and might be considerably simplified by direct reliance on well-known facts from the general theory of quotient rings; e.g., the derivation of $J + x = J + y^2$ from $J + x = (J + y)^2$ via $J+x=J^2+Jy+yJ+y^2$ seems awkward and redundant, since $(J + u)(J + v) = J + uv$ is an elementary and fundamental property of quotient rings.  The essential idea here is, of course, that the operations defined in $A$ carry over to $A/J$ provided $J$ is a (two-sided) ideal.  Thus, we can take $(J + u) + (J + v) = J + (u + v)$ and $(J + u)(J + v) = J + uv$ as given when manipulating the cosets $J + w$; and of course, $J + 0 = J$ is the additive identity in $A/J$.  And since cosets are equivalence classes under the relation $J + u \equiv J + v \Longleftrightarrow u - v \in J$, showing $u - v \in J + 0 = J$ is tantamount to proving $J + u = J + v$ in $A/J$.  We can use all these notions freely, and don't need to justify their invocation.  
The notion of a "square root" in a ring $A$, when construed as function which assigns to every $x$ a $y$ such that $x = y^2$, seems to be a rather subtle property which may reveal significant structural features of $A$ and is ideals.  And evidently, the concept generalizes and we may inquire as to when $A/J$ is possessed of an "n-th root".  It against his background I will try to present a concise version of a proof that
$A/J$ has an n-th root function if and only if, for every $x \in A$, there exists $y \in A$ with $x - y^n \in J$.
For we have, for any $J + x, J + y \in A/J$, 
$J + x = (J + y)^n \Longleftrightarrow J + x = J + y^n \Longleftrightarrow x - y^n \in J. \tag 1$
It follows from (1) that any element $J + x \in A/J$ has an n-th root $J + y$ in $A/J$ precisely when there is $J + y \in A/J$ with $x - y^n \in J$.  Since this must apply to any $A + x$ in $A/J$, the result follows.
