Answer
Infinitely many solutions $(x,y,z)$, where $x,y,z$ are integers
Work Step by Step
We are given the equation:
$x^2+y^2=z^2$
First we prove that there exists a solution of the equation.
Consider the Pythagoreic numbers:
$x=3$
$y=4$
$z=5$
$x^2+y^2=3^2+4^2=9+16=25$
$z^2=5^2=25$
Therefore $(3,4,5)$ represents a solution of the equation.
Then we prove that there are infinitely many solutions.
Note:
$x=3a$
$y=4a$
$z=5a$
We have:
$x^2+y^2=(3a)^2+(4a)^2=9a^2+16a^2=25a^2$
$z^2=(5a)^2=25a^2$
This means that $(3a,4a,5a)$ is a solution of the equation. As $a$ is any real number, it means the equation has infinitely many solutions.
Note: We could use any triplet of Pythagorean numbers (for example $(5,12,13)$, $(8,15,17)$, $(7,24,25)$).
