Does there exist a (noetherian) commutative ring $R$ and an element $a \in R$ such that $a$ is a square in every localization of $R$ but $a$ itself is not a square? 



Locally square implies square
OK, I've got it. There is no such local criterion for squareness.


Let $k$ be a field of characteristic not $2$. Take the ring of triples $(f,g,h) \in k[t]^3$, subject to the conditions that $f(1)=g(-1)$, $g(1)=h(-1)$ and $h(1)=f(-1)$. Consider the element $(t^2,t^2,t^2)$. If this were a square, its square root would have to be $(\pm t, \pm t, \pm t)$. But two of those $\pm$'s would be the same sign, and $t$ evaluated at $1$ and at $-1$ are not equal.


Now, to check that $(t^2, t^2, t^2)$ is everywhere locally a square. Geometrically, we are talking about three lines glued into a triangle. Any prime ideal has a neighborhood which is contained in the union of two neighboring lines, say the first two. On the first two lines, $(-t, t, 1)$ is a square root of $(t^2, t^2, t^2)$. 


For the suspicious, an algebraic proof. Set $u\_1=(0, (1+t)/2, (1-t)/2)$ and let $u\_2$ and $u\_3$ be the cyclic permutations thereof. We have $u\_1+u\_2+u\_3=1$ so, in any local ring, one of the $u\_i$ must be a unit. WLOG, suppose that $u\_1$ is a unit. Notice that $u\_1 (1, -t, t)^2 = u\_1 (t^2, t^2, t^2)$. So, in a local ring where $u\_1$ is a unit, $(1,t,-t)$ is a square root of $(t^2, t^2, t^2)$.


