prove that
squar root x^2+y^2 is a irrational if x and y is odd , positiv integer .
thanks
Suppose that:
$\displaystyle x, y \in \mathbb{Z}$ and that the square root of the sum of their squraes is rational, then there exist $\displaystyle a, b \in \mathbb{N}_+$ with $\displaystyle \gcd(a,b)=1$ such that:
$\displaystyle \sqrt{x^2+y^2} =\frac{a}{b}$
Then:
$\displaystyle b^2(x^2+y^2)=a^2$
Now consider any prime factor of $\displaystyle b$ and the divisibility of $\displaystyle a^2$ by this prime, and in consequence the divisibility of $\displaystyle a$ by the same prime.
This will show that if $\displaystyle \sqrt{x^2+y^2}$ is rational it is an integer and so $\displaystyle x^2+y^2$ ia a square.
CB
You need both of the responses to follow what is going on. Suppose that $\displaystyle x$ and $\displaystyle y$ are integers and that $\displaystyle \sqrt{x^2+y^2}$ is rational, then the first post shows that this is of necessity an integer and that $\displaystyle x^2+y^2$ is a square.
Then bobak shows that any square leaves remainder $\displaystyle 0$ or $\displaystyle 1$ when divided by $\displaystyle 4$, but if $\displaystyle x$ and $\displaystyle y$ are odd integers $\displaystyle x^2$ leaves remainder $\displaystyle 1$ when divided by $\displaystyle 4 $ as does $\displaystyle y^2$, so $\displaystyle x^2+y^2$ leaves a remainder of $\displaystyle 2$ and so cannot be a square. but this contradicts our assumption that $\displaystyle \sqrt{x^2+y^2}$ is rational, and si it must be irrational.
CB