If x,y,z are non-negative integers, find the number of distinct solutions for the following equation:

2x^{2}+y^{2}+z^{2}+2xy+2xz-32x-12y-20z+136=0.

Substituting (P, Q) = (x+y, x+z), and simplifying the given equation, we have:

P^2 + Q^2 – 12P – 20Q + 136 = 0

or, (P-6)^2 + (Q-10)^2 = 0

Since each of P and Q must be integers, the lhs of the above equation is zero only when:

(P-6)^2 = (Q-10)^2 = 0, so that:

(P, Q) = (6, 10)

or, (x+y, x+z) = (6, 10)……(#)

Now, x>=0, since x is nonnegative, and x cannot exceed 6 for y = 6-x will be negative if it does.

Thus, x must range from 0 to 6 inclusively, giving seven distinct pairs (y, z).

It can indeed be verified that: (x, y, z) = (0, 6, 10), (1, 5, 9), (2, 4, 8), (3, 3, 7), (4, 2, 6), (5, 1, 5) and (6, 0, 4) are the only triplets in conformity with (#),

Consequently, there are precisely seven distinct solutions to the given problem.

*Edited on ***July 14, 2008, 12:01 pm**