The square root of {y} is equal to the square root of {x²} which is also equal to 9.
\((x, \mbox{ } y) \in \mathbb{N}\)
\((x, \mbox{ } y) \in \mathbb{Z}\)
+678 So you're actually solving \(\sqrt{y} = \sqrt{x^2} = 9\) .
Ignore \( \sqrt{x^2}\), then the equation would be \(\sqrt{y} = 9\) .
Now take the left and right sides to the power of 2, so \(y = 81\)
Now ignore \(\sqrt{y}\) and take \(\sqrt{x^2}\) in the equation again.
\(\sqrt{x^2} = 9\), it applies that \(\sqrt{x^2} = x\), then \(x = 9\) ,
so
x = 9 ,
y = 81,
x and y are positive integers.
