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Math Help - Diophantine equation Legendre.

  1. #1
    Newbie individ's Avatar
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    Diophantine equation Legendre.

    Constantly there are threads with a question - how to find the solution of those equations.
    So I think that you need to draw the formula describing their solutions. At least, the formula does not hurt.
    Often there is a need to address such equations.
    So let there be this formula.
    Formula generally look like this: Do not like these formulas. But this does not mean that we should not draw them.
    To start this equation zayimemsya, well then, and others. aX^2+bXY+cY^2=jZ^2


    Solutions can be written if even a single root. \sqrt{j(a+b+c)} , \sqrt{b^2 + 4a(j-c)} , \sqrt{b^2+4c(j-a)}


    Then the solution can be written.


    X=(2j(b+2c)^2-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2 +2(b+2c)(\sqrt{j(a+b+c)}\mp{j})sp+(j\mp \sqrt{j(a+b+c)})p^2


    Y=(2j(2j-b-2a)(b+2c)-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2 +2((2j-2a-b)\sqrt{j(a+b+c)}\mp{j(b+2c)})sp+(j\mp\sqrt{j(a+b+  c)})p^2


    Z=(2j(b+2c)^2-(b^2+4c(j-a))(a+b+c\pm\sqrt{j(a+b+c)}))s^2 +2(b+2c) ( \sqrt{j(a+b+c)} \mp{j})sp + ( a + b + c \mp \sqrt{j(a+b+c)})p^2


    In the case when the root \sqrt{b^2+4c(j-a)} whole. Solutions have the form.


    X=((2j-b-2c)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2 +2(4ac+b(2j-b)\pm{(2j-b-2c)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2


    Y=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(2j-b-2a\mp\sqrt{b^2+4c(j-a)}))s^2 +2(4ac+b(2j-b)\pm{(b+2a)}\sqrt{b^2+4c(j-a)})sp+(2j-b-2a\pm\sqrt{b^2+4c(j-a)})p^2


    Z=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2 +2(4ac+b(2j-b)\pm {(b+2a)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2


    In the case when the root \sqrt{b^2+4a(j-c)} whole. Solutions have the form.


    X=(2j^2(b+2a)-j(a+b+c)(2j-2c-b\pm\sqrt{b^2+4a(j-c)}))p^2 +2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps+(2j-2c-b\mp\sqrt{b^2+4a(j-c)})s^2


    Y=(2j^2(b+2a)-j(a+b+c)(b+2a\pm\sqrt{b^2+4a(j-c)}))p^2 +2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps+(b+2a\mp\sqrt{b^2+4a(j-c)})s^2


    Z=j(a+b+c)(b+2a\mp\sqrt{b^2+4a(j-c)})p^2 +2((a+b+c)\sqrt{b^2+4a(j-c)}\mp{j(b+2a)})ps+ (b+2a\mp\sqrt{b^2+4a(j-c)})s^2


    Since these formulas are written in general terms, require a certain specificity calculations.If, after a permutation of the coefficients, no root is not an integer. You need to check whether there is an equivalent quadratic form in which, at least one root of a whole. Is usually sufficient to make the substitution X\longrightarrow{X+kY} or more Y\longrightarrow{Y+kX} In fact, this reduces to determining the existence of solutions in certain Pell's equation. Of course with such an idea can solve more complex equations. If I will not disturb anybody, slowly formula will draw. number p,s integers and set us. I understand that these formulas do not like. And when they draw - or try to ignore or delete. Formulas but there are no bad or good. They either are or they are not.


    In equation aX^2+bY^2+cZ^2=qXY+dXZ+tYZ


    a,b,c,q,d,t integer coefficients which specify the conditions of the problem.


    For a more compact notation, we introduce a replacement.


    k=(q+t)^2-4b(a+c-d)


    j=(d+t)^2-4c(a+b-q)


    n=t(2a-t-d-q)+(2b-q)(2c-d)


    Then the formula in the general form is:


    X=(2n(2c-d-t)+j(q+t-2b\pm\sqrt{k}))p^2+ 2((d+t-2c)\sqrt{k}\mp{n})ps+(2b-q-t\pm\sqrt{k})s^2


    Y=(2n(2c-d-t)+j(2(a+c-d)-q-t\pm\sqrt{k}))p^2+ 2((d+t-2c)\sqrt{k}\mp{ n })ps+(q+t+2(d-a-c)\pm\sqrt{k})s^2


    Z=(j(q+t-2b\pm\sqrt{k})-2n(2(a+b-q)-d-t))p^2+ 2((2(a+b-q)-d-t)\sqrt{k}\mp{n})ps+(2b-q-t\pm\sqrt{k})s^2


    And more.


    X=(2n(q+t-2b)+k(2c-d-t\pm\sqrt{j}))p^2+ 2((2b-q-t)\sqrt{j}\mp{n})ps+(d+t-2c\pm\sqrt{j})s^2


    Y=(2n(2(a+c-d)-q-t)+k(2c-d-t\pm\sqrt{j}))p^2+ 2((q+t+2(d-a-c))\sqrt{j}\mp{n})ps+(d+t-2c\pm\sqrt{j})s^2


    Z=(2n(q+t-2b)+k(d+t+2(q-a-b)\pm\sqrt{j}))p^2+ 2((2b-q-t)\sqrt{j}\mp{n})ps+(2(a+b-q)-d-t\pm\sqrt{j})s^2


    p,s are integers and are given us. Since formulas are written in general terms, in the case where neither the root is not an integer, it is necessary to check whether there is such an equivalent quadratic form in which at least one root of a whole. If not, then the solution in integers of the equation have not.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Diophantine equation Legendre.

    Thank you for the post, even if it is a bit on the long side. I do have one question...who is zayimemsya?

    -Dan
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  3. #3
    Newbie individ's Avatar
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    Re: Diophantine equation Legendre.

    I'm not talking English. Sometimes Google does not translate correctly.
    Hope formula understood.
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