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Conjecture re form x^2 + Mxy + y^2

  1. Feb 13, 2008 #1
    Conjecture: If x and y are coprime and M <> 2 then x^2 + Mxy +y^2 = p^2 has integral solutions only for p = a prime or for products of such primes. Also if M is positive then
    both x and y are partial solutions for X^2 -MXY + Y^2 = p^2. Thus 3*3 +3*5 +5*5 = 49 and 9 - 3*8 + 8*8 and 25 - 5*8 + 84 also equal 49
  2. jcsd
  3. Feb 13, 2008 #2
    x^2 +Mxy +y^2 == 0 mod n

    for M<2, x^2 +Mxy +y^2 = (x+y)^2 - kxy, where M = 2-k

    for M>2, x^2 +Mxy +y^2 = (x+y)^2 + kxy, where M = 2+k

    [tex](x+y)^2 == \pm kxy \ mod \ n[/tex]

    as n do not divide (x+y), then for euler

    [tex](x+y)^{\varphi(n)} == 1 \ mod \ n[/tex]

    supose n is composite ==> [tex]\varphi(n)} = 2^ah[/tex] such that a > 1 and h is odd ==>

    ==> [tex]\huge (x+y)^{2^{2^{a-1}h}} = (x+y)^{2^ah} = (x+y)^{\varphi(n)} == (\pm kxy)^{2^{a-1}h} == 1 \ mod \ n[/tex]

    call [tex](\pm kxy)^{2^{a-1}} = w[/tex], and w is obviously positive, then

    [tex]w^h == 1 \ mod \ n [/tex], an absurdum

    **then n is prime**
  4. Feb 15, 2008 #3
    But for n = 13*7 and M = 1 there is a solution, , i.e. x = 39, y = 65, how does that agree with your conclusion that n must be prime?
  5. Feb 15, 2008 #4
    Nevermind, I made a mistake in the proof, the conclusion is wrong, I tried to repair the proof without success.

    Please, define a little more clear your conjecture, what you mean saying "integral solutions"?
  6. Feb 15, 2008 #5
    Another example
    7*7 + 7*8 + 8*8 = 13^2

    19^2 + 19*80 + 80^2 = 49^2*13^2 = (7*13)^2

    Please disreegard the solution in my last post since 39 and 65 are not coprime.
    By integral solutions, I mean coprime positive integers.
  7. Feb 15, 2008 #6
    in fact n = 13*7*91, your conjecture is about p and p^2 only? or what? every composite number? I did not follow this part.
  8. Feb 15, 2008 #7
    ok, so the conjeture is: for x and y coprime, the expression is equal to some power of a prime number OR some power or a semi-prime number (a product with no more than 2 different factors)?
  9. Feb 15, 2008 #8
    if your conjecture is about only squares, then we can use the fact that a square is a number of the form 8n+1 where n is a triangular number
  10. Feb 15, 2008 #9
    How about:

    Iff M=1 and (x,y,p) are positive coprime integers, then p = a prime of the form 6n+1, or a multiple thereof, or a product of two or more primes thereof.

    Similar to Pythagorean Triples where the equation is x^2+y^2=p^2 and p = a prime of the form 4n+1...

    Added a bit later... Alternatively, for every prime p, of the form 6n+1, there is one and only one pair of positive integers (x,y) that satisfy the equation x^2+xy+y^2=p^2 (where x and y are not equal and are interchangeable).
    Last edited: Feb 16, 2008
  11. Feb 16, 2008 #10
    Whether a prime can be expressed in the form of x^2 + Mxy + y^2 where x and y are coprime depends upon the value of M. If M = 2 all integers n have upto (n-1)/2 distinct solutions But, if M = 1 and [tex] n = p_{a1}^{i1}*p_{a2}^{i2}*..p_{ah}^{ih} where p are primes, believe that there are k and only k distinct solutions where x and y are positive coprime integers (k = the sum of the powers {i1,i2,...ih} or = the number of prime factors but I dont have my work before me and I forgot) if and only if there exists solutions at all. But for M = 1 certain primes such as 3,5,11... and their products can not be expressed in this form at all. For M > 2 if x,y is a solution for n then y and (My -x) is also a solution. Note that the first few solutions for x^2 + Mxy + y^2 = n^2 occur only for n = a prime p for M > 2 since my conjecture is that if n is a product of primes then each of the primes that divide n and the the powers and products of those primes can also be expressed in the form x^2 + Mxy + y^2 for that M.
    Last edited: Feb 16, 2008
  12. Feb 19, 2008 #11
    Just calculated a interesting result
    If [tex] n = a^2 + Mab + b^2[/tex],[tex] c = b^2-a^2[/tex] and [tex]d=2ab -Mb^2[/tex]

    then [tex] n^2 = c^2 -Mcd + d^2[/tex]

    also the polarity of any 2 of M, c and d can be reversed since -1 * -1 = 1
    Last edited: Feb 19, 2008
  13. Feb 20, 2008 #12
    The above formula doesn't seem to work. Let a=b=M=1. Give n=3, c=0, d=1. Then it is not true: 3^2=0-0+1^2.

    In the case of a^2+b^2+ab, we can look at this in the form: [tex]\frac{a^3-b^3}{a-b}.[/tex] So for a not equal to b, the case for 3, we have the form: [tex]a^3\equiv{b^3}\bmod{p}[/tex] where the two integers are not the same. This shows that 3 is a divisor of p-1, and therefore p=3k+1. Since k is even, the form is 6k+1.

    This then includes the primes 7, 13, 19, 31....As suggested by Ynaugh?

    Whether there is a strictly positive integral form for the square of a prime is another matter. As ramsey2879 shows in his original entry: "Thus 3*3 +3*5 +5*5 = 49 and 9 - 3*8 + 8*8 and 25 - 5*8 + 84 also equal 49."

    So here we have the number 7, of the form 3k+1, and 7=2^2+1^2+2x1, and so there is an all positve form for the square.

    Also, I see that 13=3^2+1^1+3, and 13^2 = 8^2+7^2+56. Now, 19=3^2+2^2+6; 19^2=16^2+5^2+80.
    31=5^2+5+1; 31^2=24^2+11^2+264.

    There is, however, a key to this matter: We can decompose [tex] a^2+b^2+ab = (a-b\omega)(a-b\omega^2)[/tex], where [tex]\omega =\frac{-1+\sqrt-3}{2}[/tex]

    For example: finding the case of 37 = 4^2+3^2+12. We decompose this into:[tex] (4-3\omega)(4-3\omega^2)[/tex] then we square the first factor and simplify using the fact: [tex]1+\omega+\omega^2=0 [/tex]. We then arrive at the first factor for the square is: [tex](7-33\omega)[/tex]

    The square solution then is: [tex]37^2=33^2+7^2+33x7[/tex]

    The use of this method then makes it clear, anytime we can find the above form for the prime, of the form 6k+1, we can build on this to any higher power as well as to any product of such primes. For example: [tex]43^3=126^2+197^2+126x197.[/tex]

    However, if we are to have an all positive form then a and b must be of the different signs in the decomposition. Three is a problem in itself, [tex]3=(1-\omega)(1-\omega^2)[/tex], and there is no good form for 9, trivally, 3^2=3^2+3^3-3^2. Using the above ideas to get the form 3 times 37^2, we arrive at [tex] -26-63\omega[/tex], this produces [tex]3x37^2=26^2+73^2-26x73[/tex]
    Last edited: Feb 20, 2008
  14. Feb 20, 2008 #13
    Hey Robert Ihnot, I am a bit slow... Can you expand on this?

    Last edited: Feb 20, 2008
  15. Feb 20, 2008 #14
    O.K., actually I thought you were way ahead of me on this!

    Omega is the cube root of 1, which can be gotten from Euler's formula, here we have:
    cos(120) +isin(120). In an equation such as X^3-1 = 0. The second quadratic term, x,which is the sum of the roots, goes out. That is cos(120)+isin(120) +cos(240)+isin(240) =-1, so adding the third root cos(360)+isin(360) becomes just 0. (This may not be your question, but as a general rule the sum of the nth roots of unity is zero.)

    Multiply together[tex](a-b\omega^2)a-(a-b\omega^2)b\omega = a^2 +-ab\omega^2-ab\omega+b^2.[/tex] Now for the terms: [tex]-ab\omega-ab\omega^2=ab [/tex] So that we are left with [tex]a^2+ab+b^2[/tex] Thus what has happened has been a factoring over the complex numbers.

    Another case of this is over the Gaussian integers we have:[tex]X^2+Y^2=(X+iY)(X-iY)[/tex]
    Last edited: Feb 20, 2008
  16. Feb 20, 2008 #15
    a and b are coprime and also can not be both equal to 1, Question is 1 coprime to 1? Origionally I did not concider this exception. Thanks for pointing it out.
  17. Feb 21, 2008 #16
    LOL I am just trying to understand a picture and I did not mean to hi-jack the thread...

    Thank you. That helped.
  18. Feb 21, 2008 #17
    1s are coprime? Well, the important matter is that if p, a prime, =a^2+b^2+ab, it is obvious that a and b could not have a common factor greater than 1. And the case for 3 is that a=b=1.
    Last edited: Feb 21, 2008
  19. Feb 21, 2008 #18
    Yes but c^2 -cd + d^2 <> 9 for c and d as integers. Also, the important matter is that a <> b. How about n is not prime if the greatest commom divisor of c and d <> 1 since the greatest common divisor of c and d divides n If gcd(c,d) = 1 then n^2 = c^2 -Mcd + d^2?
    Last edited: Feb 21, 2008
  20. Feb 21, 2008 #19
    You can work on that!
  21. Feb 21, 2008 #20
    Now that I had more time to consider this I did make a typo in the formula for d it should have been 2ab +mb^2 so my form reduces to 0^2 -0*3 + 3^2 = 3^2

    Note That If you square [tex]a-b\omega[/tex] you get [tex]a^2 - 2ab\omega + b^2\omega^2 = a^2-b^2 - (2ab+b^2)\omega[/tex] so c = a^2-b^2 and d = (2ab+b^2).

    Thus c^2 +cd + d^2 = [-c]^2 -[-c]d + d^2 = n^2 which is my result.

    Also this proves that gcd(c,d) divides n since each term on the left of my equation is divisable by the square thereof.

    A interesting portion of my conjecture is that if n is composite then each factor of n must be of the form a^2 + Mab + b^2 but as you said in the last post M = 13 is a problem since a = 0 and b = 3 is a trival result, which I think is removed by the requirement that gcd(c,d) must equal 1. However, your showing that [tex]3x37^2=26^2+73^2-26x73[/tex] needs to be considered. There [tex]c = 73^2 - 26^2 = 4653[/tex] Since 3|4653 then the gcd(c,d) at least equals 3. Thus this does not contradict my conjecture that all n of the general form where gcd(c,d) = 1 can be expressed as a product of primes each of which is of the form where the gcd(c,d) = 1.

    PS Do you have any suggestion of how to extend this to the general form where |M|>2?

    If my conjecture is true this result can find large primes by keeping a = 1, b = 2 and choosing a large M not divisible by 3 such that the gcd(c,d) = 1.
    Last edited: Feb 21, 2008
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