MathSolver94 wrote: Find the number of triples of nonnegative integers $(x,y,z)$ such that \[x^2+2xy+y^2-z^2=9.\] $(x+y)^2-z^2=9$ $a^2-b^2=9$ has only $a-b=1,a+b=9$ or $a=3,b=0$ and so $10$ So the edit: miscount always with such problems i see

SCP wrote: MathSolver94 wrote: Find the number of triples of nonnegative integers $(x,y,z)$ such that \[x^2+2xy+y^2-z^2=9.\] $(x+y)^2-z^2=9$ $a^2-b^2=9$ has only $a-b=1,a+b=9$ or $a=3,b=0$ and so $9$ In fact $10$ such triples : $(x+y,z)=(3,0)$ gives $(x,y,z)\in\{(0,3,0),(1,2,0),(2,1,0),(3,0,0)\}$ $(x+y,z)=(5,4)$ gives $(x,y,z)\in\{(0,5,4),(1,4,4),(2,3,4),(3,2,4),(4,1,4),(5,0,4)\}$

The problem state that $(x, y, z)$ is nonnegative, the answer is 4 edit: sorry for the misinterpret

Nonnegative, as its common meaning in English language, means larger than or equal to zero (positive means larger than zero). Thus $0$ is a nonnegative integer.