14) How many distinct ordered triples $(x, y, z)$ formed by positive integers (strictly greater than zero) are there such that

$$
x^{2}+2 x y+y^{2}-z^{2}=9 ?
$$

(A) None,
(B) two,
(C) three,
(D) four,
(E) more than six.
14. The answer is $(\mathrm{D})$.

From the relation contained in the problem, we derive:

$$
(x+y)^{2}-z^{2}=9
$$

or

$$
(x+y+z)(x+y-z)=9 .
$$

Therefore, $(x+y+z)$ and $(x+y-z)$ must be divisors of 9. In particular, since $x, y$, and $z$ are positive, $(x+y+z)$ is positive, and so must $(x+y-z)$ be. The number 9 can only be factored in two ways as a product of positive integers: $9=9 \times 1, 9=3 \times 3$. Since $(x+y+z)>(x+y-z)$ (given that $z>0$), the only possibility is

$$
\left\{\begin{array}{l}
x+y+z=9 \\
x+y-z=1
\end{array}\right.
$$

This system leads to $z=4$ and $x+y=5$, and subsequently to the triples $(1,4,4),(2,3,4),(3,2,4)$, $(4,1,4)$. In total, we have four solutions.