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

  • Context: MHB 
  • Thread starter Thread starter evinda
  • Start date Start date
  • Tags Tags
    Square
Click For Summary

Discussion Overview

The discussion revolves around the problem of showing that the expression \(x^2 + y^2\) is not a perfect square under the condition that \((x, 3) = (y, 3) = 1\). Participants explore various mathematical approaches, including modular arithmetic, to analyze the properties of the expression.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest using modular arithmetic, specifically modulo 3, to analyze the values of \(x^2\) and \(y^2\).
  • It is noted that \(x\) can take the forms \(3k + 1\) or \(3k + 2\), leading to different values of \(x^2 \mod 3\).
  • One participant questions whether \(x\) could also be \(3k\), which would change the gcd condition.
  • There is a consensus that \(x^2\) and \(y^2\) both yield a remainder of 1 when divided by 3, leading to the conclusion that \(x^2 + y^2 \mod 3 = 2\).
  • Participants discuss the implications of \(x^2 + y^2\) being equal to \(b^2\) for some integer \(b\), analyzing the possible values of \(b^2 \mod 3\).
  • It is proposed that since \(x^2 + y^2 \mod 3\) can only equal 2, it cannot be a perfect square, although this conclusion is not universally accepted as definitive.

Areas of Agreement / Disagreement

While there is some agreement on the modular analysis leading to the conclusion that \(x^2 + y^2\) cannot be a perfect square, the discussion contains elements of uncertainty and differing interpretations regarding the implications of the gcd condition and the validity of the conclusions drawn.

Contextual Notes

Participants express uncertainty about the implications of assuming \(x = 3k\) and its effect on the gcd condition. The discussion also reflects a reliance on modular arithmetic without fully resolving all assumptions or steps in the reasoning.

evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
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:
 
Mathematics news on Phys.org
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! :)
 

Similar threads

MHB Condition for A Quartic Equation to have a Real Root
  • · Replies 2 ·
Replies
2
Views
1K
MHB Mil.navy.01 completing the square
  • · Replies 2 ·
Replies
2
Views
1K
MHB Prove No Integers Solve $ax^3+bx^2+cx+d=1$ for x=19,2 for x=62
  • · Replies 3 ·
Replies
3
Views
2K
MHB Polynomial Challenge: Show Real Roots >1 Exist
  • · Replies 1 ·
Replies
1
Views
1K
High School A symmetric problem has an asymmetric solution - why?
  • · Replies 21 ·
Replies
21
Views
2K
Undergrad Quadratic, cubic, quartic, quintic equations
  • · Replies 16 ·
Replies
16
Views
5K
MHB How Can You Rewrite an Equation to Express y as a Function of x?
  • · Replies 3 ·
Replies
3
Views
1K
MHB Prove Divisibility: $(x-y)^2+(y-z)^2+(z-x)^2=xyz$ yields $x^3+y^3+z^3$
  • · Replies 1 ·
Replies
1
Views
2K
MHB Solving System of Equalities: x^2-y\sqrt xy & y^2-x\sqrt xy =3
  • · Replies 2 ·
Replies
2
Views
1K
MHB What is the graph of y = sin^2 x + cos^2 x?
  • · Replies 1 ·
Replies
1
Views
1K