1. ## prove

prove that

squar root x^2+y^2 is a irrational if x and y is odd , positiv integer .

thanks

2. Originally Posted by hoger
prove that

squar root x^2+y^2 is a irrational if x and y is odd , positiv integer .

thanks
Suppose that:

$x, y \in \mathbb{Z}$ and that the square root of the sum of their squraes is rational, then there exist $a, b \in \mathbb{N}_+$ with $\gcd(a,b)=1$ such that:

$\sqrt{x^2+y^2} =\frac{a}{b}$

Then:

$b^2(x^2+y^2)=a^2$

Now consider any prime factor of $b$ and the divisibility of $a^2$ by this prime, and in consequence the divisibility of $a$ by the same prime.

This will show that if $\sqrt{x^2+y^2}$ is rational it is an integer and so $x^2+y^2$ ia a square.

CB

3. You can also consider quadratic residues mod 4. It is easy to show that all squares leave a remainder of 1 or 0 when divided by 4.
$x^2 + y^2$ leaves a reminder of 2 if both x and y are odd so it cannot be a perfect square.

Bobak

4. Originally Posted by bobak
You can also consider quadratic residues mod 4. It is easy to show that all squares leave a remainder of 1 or 0 when divided by 4.
$x^2 + y^2$ leaves a reminder of 2 if both x and y are odd so it cannot be a perfect square.

Bobak

hi
can you explain more for me .

thank you

5. Originally Posted by hoger
hi
can you explain more for me .

thank you
You need both of the responses to follow what is going on. Suppose that $x$ and $y$ are integers and that $\sqrt{x^2+y^2}$ is rational, then the first post shows that this is of necessity an integer and that $x^2+y^2$ is a square.

Then bobak shows that any square leaves remainder $0$ or $1$ when divided by $4$, but if $x$ and $y$ are odd integers $x^2$ leaves remainder $1$ when divided by $4$ as does $y^2$, so $x^2+y^2$ leaves a remainder of $2$ and so cannot be a square. but this contradicts our assumption that $\sqrt{x^2+y^2}$ is rational, and si it must be irrational.

CB