14. How many ordered pairs $(x, y)$ of integers satisfy the equation $y^{4}-8 y^{2}+7=8 x^{2}-2 x^{2} y^{2}-x^{4} ?$
14. The answer is 4. The proposed equation is equivalent to:

$$
\left(x^{2}+y^{2}-1\right)\left(x^{2}+y^{2}-7\right)=0 \text {, }
$$

whose solutions are given by the pairs $(x, y)$ of integers that satisfy $x^{2}+y^{2}=1$ or $x^{2}+y^{2}=7$. However, the latter equation is impossible, since an integer that is the sum of two squares cannot have a remainder of 3 when divided by 4. It follows that the only solutions are $( \pm 1,0)$ and $(0, \pm 1)$. (Alfieri)