Number theory: Binary Quadratic Forms

In summary, the conversation discusses how to prove that the equation 2x^2+2xy+3y^2=5 has no integer solutions. The conversation includes discussions on completing the square and considering different cases, ultimately concluding that the equation is impossible as it would require x, y, and x+y to be divisible by 5, which is not possible.
  • #1
kingwinner
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P.S. I'm not sure where to post this question, in particular I can't find a number theory forum on the coursework section for textbook problems. Please move this thread to the appropriate forum if this is not where it should belong to. Thanks!
 
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  • #2
nt11.JPG
 
  • #3
All I can tell you is that if x^2 + 5y^2 ==0 Mod p. Then it follows that (x/y)^2 ==-5 Mod p. But looking at that Mod 4, we have the form x^2 + y^2 ==1 Mod 4, because they can not be both even or odd, otherwise p would be divisible by 2.
The form 4k+1 has -1 as a quadratic residue. So we have a solution to u^2 ==5 Mod p. From the form (p/5)=1, we have the solution v^2 ==+/-1 Mod 5. To fulfill both these cases, we have 20k+1, 20k+9.

By looking at the various cases we can see that in the form 2x^2+2xy +3Y^2 =p, we know that y must be odd, and that for either x is odd or even we arrive at p== 3y^2==3 Mod 4. Thus -1 is not a quadratic residue Mod p. As you have written (p/5) = -1. So p is not a quadratic residue Mod 5. Thus p is of the form p= 5k+/- 2 Mod 5. We have the result p = 20k+ 3 or 7.

But you may already know this.
.
 
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  • #4
kingwinner said:
P.S. I'm not sure where to post this question, in particular I can't find a number theory forum on the coursework section for textbook problems. Please move this thread to the appropriate forum if this is not where it should belong to. Thanks!

Questions about number theory may be posted in the Calculus & Beyond section in the coursework forum.

Petek
 
  • #5
So for part c, I think we have to show that the case p=5 is impossible.

In short, my question just translates to the following:
How can we prove that [tex] 2x^{2}+2xy+3y^{2} =5 [/tex] has no integer solutions?

Thanks!
 
  • #6
You might get that by brute force. We know that both x and y can not be positive since even x=y=1 will sum to 7. It doesn't matter which is negative, but we rewrite the equation as 2x^2-2xy+3y^2. Y must be odd and if as large as 3, we have 2x^2-6x=-22. This gives x^2-3x = 11. Which gives x(x-3) =11, which is a prime and so this is impossible.

Possibly, by proceeding in this way, we can figure this thing out.

Once we know a few facts from the above, we can look at the matter as a congruence modulo 4. In this case y and all odd squares are congruent to 1. So the matter reduces to 2X^2 +2xy==-2 Mod 4, and then to x^2+xy ==1 Mod 2. So x must be odd, x^2==1 Mod 2, and thus xy==0 Mod 2, which is impossible. Maybe that will work!
 
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  • #7
Claim: [tex]2x^{2}+2xy+3y^{2}=5[/tex] has no integer solutions

If we complete the square, would that help?
[tex](x+y)^{2}+x^{2}+2y^{2} = 5[/tex]

Is it possible to eliminate some cases from here and show the remaining cases do not give 5? If so, how?
 
  • #8
kingwinner said:
Claim: [tex]2x^{2}+2xy+3y^{2}=5[/tex] has no integer solutions

If we complete the square, would that help?
[tex](x+y)^{2}+x^{2}+2y^{2} = 5[/tex]

Is it possible to eliminate some cases from here and show the remaining cases do not give 5? If so, how?


(Post by originator, so I assume it was a question rather than a hint.)

The answer then drops out. You can't have [itex]x=y=0(5)[/itex] otherwise [itex]5^2\mid[/itex] LHS but [itex]5^2\nmid[/itex] RHS.

So [itex]x\neq 0(5)[/itex], [itex]y\neq 0(5)[/itex] and [itex]x+y\neq 0(5)[/itex].

So [itex](x+y)^{2}+x^{2}+2y^{2}\geq 7[/itex].
 

1. What is number theory?

Number theory is a branch of mathematics that deals with the properties and relationships of integers.

2. What are binary quadratic forms?

Binary quadratic forms are algebraic expressions of the form ax^2 + bxy + cy^2, where a, b, and c are integers. These expressions are used to study the properties of quadratic equations.

3. What is the significance of binary quadratic forms in number theory?

Binary quadratic forms play a crucial role in number theory, as they can be used to represent and solve various problems related to integers, such as finding solutions to Diophantine equations and studying the distribution of primes.

4. How are binary quadratic forms related to quadratic residues?

In number theory, a quadratic residue is an integer that is congruent to a perfect square modulo a given integer. Binary quadratic forms can be used to determine the existence and characteristics of quadratic residues for a given integer.

5. What are some modern applications of number theory and binary quadratic forms?

Number theory and binary quadratic forms have many practical applications, such as in cryptography, coding theory, and computer science. They are also used in the design and analysis of algorithms for solving mathematical problems.

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