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Number theory: Binary Quadratic Forms

  1. Mar 21, 2010 #1

    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!
    Last edited: Mar 21, 2010
  2. jcsd
  3. Mar 21, 2010 #2
  4. Mar 22, 2010 #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.
    Last edited: Mar 22, 2010
  5. Mar 22, 2010 #4
    Questions about number theory may be posted in the Calculus & Beyond section in the coursework forum.

  6. Mar 22, 2010 #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?

  7. Mar 22, 2010 #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!
    Last edited: Mar 22, 2010
  8. Mar 22, 2010 #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?
  9. Apr 3, 2010 #8

    (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].
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