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

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SUMMARY

The discussion centers on proving that the expression \(x^2 + y^2\) cannot be a perfect square when \( (x, 3) = (y, 3) = 1 \). Participants explore various approaches, ultimately concluding that by analyzing \(x^2 + y^2 \mod 3\), where both \(x^2\) and \(y^2\) yield a remainder of 1, the sum \(x^2 + y^2\) results in a remainder of 2. This contradicts the possible remainders of perfect squares modulo 3, which can only be 0 or 1. Therefore, \(x^2 + y^2\) cannot be a perfect square.

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evinda
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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?? :confused:
 
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evinda said:
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?? :confused:

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?
 
I like Serena said:
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 $x=3k$ ?? :confused:

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?? :confused:
 
evinda said:
Couldn't it also be $x=3k$ ?? :confused:

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?? :confused:

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 $x^2+y^2$?
 
I like Serena said:
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 $3$! :o
I like Serena said:
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 $x^2+y^2\text{ mod } 3=2$ ??
 
evinda said:
Oh yes,you are right!It would be equal to $3$! :o

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

Yep and yep! :)
 
I like Serena said:
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??
 
evinda said:
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.
 
I like Serena said:
Correct.

Great...Thank you very much! :)
 

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