Which integers are the areas of squares with vertices in the 3d integer lattice? For which integers $n$ does there exist a square of area $n$ with vertices in the 3d integer lattice $\mathbb{Z}^3$? 
A sufficient condition is that $n$ is a square or the sum of two squares, and I have verified that the condition is also necessary when $n < 10^5$.
Edit: This question was posed by James Tanton on Twitter. I thought it was very interesting, so I took the liberty of posting it here.
Edit 2: I have extended the search to $n < 10^6$ without finding any counterexamples.

With Qiaochu's observation, one can find a general parametric solution to this problem, starting from:
$a^2 + b^2 + c^2 = d^2 + e^2 + F^2$
$a d + b e + c F = 0$
Start by forgetting about integers for now, and dividing both equations by $c^2$, and considering $a, b, d, e, F$ as rational. In other words, in effect take $c = 1$.
Then plugging $- F = a d + b e$ into the first gives a result equivalent to:
$(a^2 + 1) d^2 + 2 a b d e + (b^2 + 1) e^2 = a^2 + b^2 + 1$
Multiplying throughout by $a^2 + 1$, this can be expressed in the form:
$((a^2 + 1) d + a b e)^2 = (a^2 + 1 - e^2) (a^2 + b^2 + 1)$
Letting:
$(a^2 + 1) d + a b e = (a^2 + b^2 + 1) f$
this becomes:
$a^2 + 1 - e^2 = (a^2 + b^2 + 1) f^2$
or equivalently:
$(a^2 + 1) (1 - f^2) = e^2 + (b f)^2$
It isn't hard to prove that this implies the existence of rational $g, h$ with:
$1 - f^2 = g^2 + h^2$
whence by composition:
$e^2 + (b f)^2 = (a g + h)^2 + (a h - g)^2$
This in turn implies the existence of rational $u, v$ with:
$u^2 + v^2 = 1$
such that again by composition:
$e = u (a g + h) + v (a h - g)$
$b f = v (a g + h) - u (a h - g)$
Now $f^2 + g^2 + h^2 = 1$ has general solution:
$f = \frac{p^2 + q^2 - 1}{p^2 + q^2 + 1}$
$g = \frac{2 p}{p^2 + q^2 + 1}$
$h = \frac{2 q}{p^2 + q^2 + 1}$
and $u^2 + v^2 = 1$ of course has general solution:
$u = \frac{r^2 - 1}{r^2 + 1}$
$v = \frac{2 r}{r^2 + 1}$
So plugging these two solutions into the preceding equations for $e$ and $b f$ expresses $b$ and $e$ in terms of $a, p, q, r$.
Also plugging these two solutions, and $b$ and $e$ just obtained, into the equation (near 
the start) where $f$ was introduced, expresses $d$ in terms of $a, p, q, r$.
Finally, $F$ follows from $- F = a d + b e$.
Then simply homogenize to obtain equations giving all integer solutions.

I later realized that the above, although correct, is suboptimal.
Solving $a^2 _+ b^2 + 1 = d^2 + e^2 + (a d - b e)^2$, as we did, by expressing $b, d, e$ in terms of $a$ and parameters should require only 2 parameters instead of our 3.
This mystery can be resolved by observing that $u^2 + v^2 (= 1)$ can be absorbed by composition into $g^2 + h^2$. So with a pair of new parameters $G, H$, with $1 - f^2 = G^2 + H^2$ we can conclude:
$e = a G + H$
$b f = a H - G$
and $b, d, e, F$ can each now be expressed in terms of $a, G, H$
