, 2012

The answer is $518400$.

We claim that $3^4 \cdot 4^4 \cdot 5^2$ divides $xyzuvw$.

Firstly, observe that $3$ divides $x$ or $3$ divides $y$, since otherwise
$$u^2 = x^2 + y^2 \equiv 1 + 1 = 2 \pmod{3},$$
which is impossible. Similarly, $3$ divides $x$ or $z$, and $3$ divides $y$ or $z$. Thus, two of $x$, $y$, $z$ are divisible by $3$, say $x$ and $y$. Then we have $3$ divides $u$, and the first equation becomes $\left(\frac{x}{3}\right)^2 + \left(\frac{y}{3}\right)^2 = \left(\frac{u}{3}\right)^2$. Again, $3$ divides one of $\frac{x}{3}$ and $\frac{y}{3}$. Thus, we have $3^4$ divides $xyu$. In general, we must have $3^4$ divides $xyzuvw$.

Secondly, observe that $4$ divides $x$ or $4$ divides $y$, since otherwise
$$u^2 = x^2 + y^2 \equiv 1 + 1 = 2 \pmod{4},$$
which is impossible. As above, this implies $4^4$ divides $xyzuvw$.

Thirdly, observe that $5$ divides $x$, $y$ or $u$, since otherwise
$$x^2 + y^2 \equiv \pm 1 \pm 1 = \pm 2 \not\equiv \pm 1 \equiv u^2 \pmod{5}.$$
In the same way, $5$ divides one of $x$, $z$, $v$, $5$ divides one of $y$, $z$, $w$. It is not hard to see $5$ divides at least two of $x$, $y$, $z$, $u$, $v$, $w$, and hence $5^2$ divides $xyzuvw$.

Combining these, we obtain $3^4 \cdot 4^4 \cdot 5^2 = 518400$ divides $xyzuvw$.