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I would like to find a parametric solution for the following diophantine equation:

$-4 (1-a_1^2)(1-a_2^2) + (1+a_3^2 -a_1^2 -a_2^2)^2 = a_4^2$

Does such a solution exist? How does one go about solving such questions systematically?

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  • $\begingroup$ Could you write this equation in this form? $q^2=(z^2+d^2-x^2-y^2)^2-4(d^2-x^2)(d^2-y^2)$ And look for a solution in integers. Because this equation has a solution. $\endgroup$
    – individ
    Commented May 30, 2014 at 17:19
  • $\begingroup$ Yes indeed, this amounts to a rescaling of my variables. Where can I find a solution to that equation? $\endgroup$
    – Johannes
    Commented May 30, 2014 at 17:23
  • $\begingroup$ Nowhere. It is necessary to solve this equation. $\endgroup$
    – individ
    Commented May 30, 2014 at 17:24
  • $\begingroup$ Ok, I see. Are there somewhat systematic methods for solving equations like this? Perhaps it is better to start with a simpler version of this. E.g., the following eq. is also relevant for me, $(1 - b2^2) (1 - b3^2) - 4 (1 - b1^2) = 0$. It has one variable less. $\endgroup$
    – Johannes
    Commented Jun 2, 2014 at 13:29
  • $\begingroup$ For such equations the more unknown so easily solved. It is better not to reduce the number of unknowns. And what is the equation you need? $\endgroup$
    – individ
    Commented Jun 2, 2014 at 14:00

1 Answer 1

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Equation:

$q^2=(z^2+d^2-x^2-y^2)^2-4(d^2-x^2)(d^2-y^2)$

Has the solution:

$x=(p-s)sa^2+(p^2-ps+s^2)t^2-psk^2$

$y=-(p-s)sa^2+(p^2-ps+s^2)t^2-psk^2+2psak-2(p^2-ps+s^2)at$

$d=(p-s)sa^2-(p^2-ps+s^2)t^2-psk^2-2(p-s)sat+2pskt$

$z=(p-s)sa^2+(p^2-ps+s^2)t^2+psk^2-2(p^2-ps+s^2)kt-2(p-s)sak$

$q=4((p^2-ps+s^2)((p^2-s^2)k-(p^2-2ps)a)t^3+ps((p^2-s^2)t+as^2)k^3+$

$+s((p^3-3sp^2+2ps^2)t+k(p-s)s^2)a^3-(p^4-s^4)t^2k^2+p(p^3-4sp^2+6ps^2-4s^3)t^2a^2-$

$-(2p-s)s^3a^2k^2+(2p^3-3sp^2-ps^2+s^3)sakt^2-(p^3-2ps^2+2s^3)satk^2-$

$-(p^3-3sp^2+ps^2-s^3)skta^2)$

$p,s,a,t,k$ - Any integers.

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