, 2009

Solution:
The first equation always has solutions, namely the triples $\{x, x+1, x(x+1)-c\}$ for all $x \in \mathbb{N}$. Indeed,
$$\left(x^{2}-c\right)\left((x+1)^{2}-c\right)=x^{2}(x+1)^{2}-2c\left(x^{2}+(x+1)^{2}\right)+c^{2}=(x(x+1)-c)^{2}-c.$$

For the second equation, we try $z=|xy-c|$. We need
$$\left(x^{2}+c\right)\left(y^{2}-c\right)=(xy-c)^{2}$$
or
$$x^{2}y^{2}+c\left(y^{2}-x^{2}\right)-c^{2}=x^{2}y^{2}-2xyc+c^{2}$$
Cancelling the common terms we get
$$c\left(x^{2}-y^{2}+2xy\right)=2c^{2}$$
or
$$c=\frac{x^{2}-y^{2}+2xy}{2}$$
Therefore, all $c$ of this form will work. This expression is a positive integer if $x$ and $y$ have the same parity, and it clearly takes infinitely many positive values. We only need to check $z \neq 0$, i.e. $c \neq xy$, which is true for $x \neq y$. For example, one can take
$$y=x-2$$
and
$$z=\frac{x^{2}-(x-2)^{2}+2x(x-2)}{2}=x^{2}-2.$$
Thus, $\{(x, x-2,2x-2)\}$ is a solution for $c=x^{2}-2$.