Dealing with Integers!

When x=1,

It becomes z+1 = y^2.

So, z can be {3,8,,15,24,...........}.

So, there are infinite pairs of (x,y,z).

Even at x=2, we can have infinite values as z can take such a value that x(x+z) will be a perfect square, like z={6,16,30,......}.

So, for any value of x, z can take such a value so that x(x+z) will be a perfect square.

$x(x+z) = y^2 \text{ /Let z = 0} \\ \Rightarrow x^2 = y^2 \\ \Rightarrow x = \pm y \\ \text{There are infinite pairs of (x, y) satisfying the above equation.}$

There are very few restrictions in this problem, so we think to start by substituting a value for one of the variables and investigate a special case.

The simplest idea is to let $x=1,$ which gives $1+z = y^2.$ Of course, this equation has infinite solutions since we can take any positive integer $y\ge 2$ and then $z=y^2 -1$ is a positive integer as well. For example, we could have $(x,y,z) = (1,3,8),$ so $1\cdot (1+8) = 3^2.$

Since there are infinite solutions of the form $\left(1,y,y^2-1\right),$ the equation has infinite solutions!

Looking for more of a challenge? Check out this post by Daniel Liu which explores an extension of this problem!

Even when x=1, there are an infinite number of tuples that satisfy the equation, so the correct answer is INFINITY.

when x=1, there are an infinite number of tuples that satisfy the equation, so the correct answer is INFINITY.

x(x+z) = y^2, implies that , (x)(z)=(y+x)(y-x) , thus we get a condition , when x = (y-x) , then z=( y+x), which follows that , 2(x) = y and 3(x) = z . Hence, the relations- y=2(x) and z = 3(x) satisfy the above equation.Now, taking any positive integer as x , we can get the ordered pairs of positive integers (x,y,z), which satisfies x(x+z) = y^2, and as there are infinitely many positive integers , then, there exist infinitely many (x,y,z).

Even at x=1, there are infinite such pairs, so done.