Show that x^2+y^2 is not a square

[evinda]

Hello! :)

Let $(x,3)=(y,3)=1$.Show that $x^2+y^2$ is not a square.How can I do this?

I thought that I could use these relations:
$$ax+3b=1$$
$$yc+3d=1$$
But,using them I found: $x^2+y^2=\frac{cx(1-3b)+ay(1-3d)}{ac}$.. I think that we can't conclude it from this..or am I wrong??

 

[I like Serena]

Hello! :)

Let $(x,3)=(y,3)=1$.Show that $x^2+y^2$ is not a square.How can I do this?

I thought that I could use these relations:
$$ax+3b=1$$
$$yc+3d=1$$
But,using them I found: $x^2+y^2=\frac{cx(1-3b)+ay(1-3d)}{ac}$.. I think that we can't conclude it from this..or am I wrong??

Hi! :D

That's looking a bit complicated.

Let's try a different approach. Suppose we calculate modulo 3...

Then $x = 3k+1$ or $x=3k+2$.
What is $x^2 \text{ mod } 3$ in those 2 cases?

Following up, what can $x^2 + y^2 \text{ mod } 3$ be?

And suppose it would be a square, can that match?

 

[evinda]

Hi! :D

That's looking a bit complicated.

Let's try a different approach. Suppose we calculate modulo 3...

Then $x = 3k+1$ or $x=3k+2$.
What is $x^2 \text{ mod } 3$ in those 2 cases?

Following up, what can $x^2 + y^2 \text{ mod } 3$ be?

And suppose it would be a square, can that match?

Couldn't it also be $x=3k$ ??

I found that: $x^2 \text{ mod } 3=0$ for $r=0$,$x^2 \text{ mod } 3=1$ for $r=1$ and $x^2 \text{ mod } 3=1$ for $r=1$..But how can I continue??

 

[I like Serena]

Couldn't it also be $x=3k$ ??

I'm assuming that with $(x,3)=1$, you mean that the greatest common denominator of x and 3 is 1. Suppose x=3k, what would the gcd be?

I found that: $x^2 \text{ mod } 3=0$ for $r=0$,$x^2 \text{ mod } 3=1$ for $r=1$ and $x^2 \text{ mod } 3=1$ for $r=1$..But how can I continue??

Disregarding your result for $x\text{ mod } 3=0$, you can see that $x^2$ always has remainder 1 when divided by 3.
What would the remainder be of $x^2+y^2$?

 

[evinda]

I'm assuming that with $(x,3)=1$, you mean that the greatest common denominator of x and 3 is 1. Suppose x=3k, what would the gcd be?

Oh yes,you are right!It would be equal to $3$!

Disregarding your result for $x\text{ mod } 3=0$, you can see that $x^2$ always has remainder 1 when divided by 3.
What would the remainder be of $x^2+y^2$?

$y^2$ has also always remainder 1 when divided by 3.So,would $x^2+y^2\text{ mod } 3=2$ ??

 

[I like Serena]

Oh yes,you are right!It would be equal to $3$!

$y^2$ has also always remainder 1 when divided by 3.So,would $x^2+y^2\text{ mod } 3=2$ ??

Yep and yep! :)

 

[evinda]

Yep and yep! :)

Then we suppose that it is a square,so it is like that: $x^2+y^2=b^2, b \in \mathbb{Z}$.
$b$ mod $3$ can take the following values: $0,1,2$,right?? So,the possible values for $b^2$ mod $3$ are $0,1,1$.
But $x^2+y^2$ mod $3$ can only be equal to $2$,so we conclude that it can't be $x^2+y^2=b^2, b \in \mathbb{Z}$.
Is it right or am I wrong??

 

[I like Serena]

Then we suppose that it is a square,so it is like that: $x^2+y^2=b^2, b \in \mathbb{Z}$.
$b$ mod $3$ can take the following values: $0,1,2$,right?? So,the possible values for $b^2$ mod $3$ are $0,1,1$.
But $x^2+y^2$ mod $3$ can only be equal to $2$,so we conclude that it can't be $x^2+y^2=b^2, b \in \mathbb{Z}$.
Is it right or am I wrong??

Correct.

 

[evinda]

Correct.

Great...Thank you very much! :)