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Interested in the solution in general Diophantine equations of the form:

$X^3+Y^3+Z^3=3XYZ+q$

$q$ - what some integer.

Solutions similar equations can be written.

Since this equation is easy, as it is quite symmetrical.

$X^3+Y^3+Z^3-3XYZ=R^3$

Such a solution can write.

$X=sp(p+s)$

$Y=s(2p^2+s^2)$

$Z=p(p^2+2s^2)$

$R=p^3+s^3$

And solutions can be written:

$X=s(9p^2+9ps+10s^2)$

$Y=s(6p^2+12ps+7s^2)$

$Z=3p^3+3p^2s+15ps^2+7s^3$

$R=3(p+2s)(p^2-ps+s^2)$

Whether there are any thoughts how to solve this equation?

At first I thought to use for solving Pell's equation, but I think that you can do without.

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2 Answers 2

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You can use the following identity:

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$$

Fixed integer q has a finite number of factorial expansions. So we get a finite number of systems of two equations:

$$x+y+z = a$$ $$x^2+y^2+z^2-xy-xz-yz = b$$

for all integers $a$ and $b$ such that $ab=q$. Excluding the variable z we obtain the quadratic equation in x and y.

CALCULATIONS:

Let $u = x + y$ and $v=x y$. Then we have $z = a - u$ and $$x^2+y^2+z^2-xy-xz-yz = u^2 - 2v + (a-u)^2 - v - (a-u)u =$$ $$u^2 - 3v + a^2 - 2au + u^2 - au +u^2 = 3u^2 - 3au - 3v + a^2$$

Thus, we have a equation in integers u and v: $$ 3(u^2 - au) - 3v = b - a^2$$ It`s something like Pell's equation as you said. But it is more simple and we can express v from u:

$$ 3v = 3(u^2 - au) - (b - a^2)$$

Next step. Let suppose that (x,y) is an integer solution of original equation then because $u = x + y$ and $v=x y$ there is integer m such that $u^2-4v=m^2$ (this is a discriminant of a square equation $(\lambda - x)(\lambda - y) = \lambda^2-u\lambda+v=0$).

So $4v = u^2 - m^2$ and we have the following transformations of the equation: $$ 3v = 3(u^2 - au) - (b - a^2)$$ $$12(u^2 - au)-3*4v = 4(b-a^2)$$ $$12(u^2 - au)-3(u^2-m^2)=4(b-a^2)$$ $$9u^2 - 12au + 3m^2=4(b-a^2)$$ $$9u^2 - 12au + 4a^2 + 3m^2=4b$$ $$(3u-2a)^2 + 3m^2=4b$$

Thus, we have that the number of solution for each fixed a and b is finite! So for each parameter q we can find all solutions (x,y,z) of the original equation $x^3+y^3+z^3 = 3xyz + q$ by the following formulas:

$$x=\frac{u-m}{2}$$ $$y=\frac{u+m}{2}$$ $$z=a-u$$

where u and m form a solution of the equation:

$$(3u-2a)^2 + 3m^2=4b$$

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  • $\begingroup$ I talked about it, and in his post. Solving this system of equations I come to Pell's equation. Thanks anyway. I'll think about how to do without it. $\endgroup$
    – individ
    May 27, 2014 at 17:01
  • $\begingroup$ You can make further restrictions on $b$: given $(x-y)^2+(y-z)^2+(z-x)^2=2b$ we know that $b$ is not of the form $2^{2k+1}(8m+7)$. $\endgroup$ May 27, 2014 at 17:07
  • $\begingroup$ After writing this you actually say that for a given number $q$ - number of solutions of course, but it is not. The number of solutions is infinite. And in general. What a way to bring solutions - without writing formulas of their decisions? $\endgroup$
    – individ
    May 27, 2014 at 17:25
  • $\begingroup$ I am calculate and as you said, individ, get something like Pell's equation :) $\endgroup$
    – DenisMath
    May 27, 2014 at 17:38
  • $\begingroup$ Good for you. One does not solve the equation changed to another. In number theory, such manner of calculation. So solve the equation, so it is not addressed. The equation is solved - this is when the formula is written receipt of the decision without further action. $\endgroup$
    – individ
    May 27, 2014 at 17:51
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Strangely enough, the solution is finite.

for the equation:

$X^3+Y^3+Z^3-3XYZ=q=ab$

If it is possible to decompose the coefficient as follows:

$4b=k^2+3t^2$

Then the solutions are of the form:

$X=\frac{1}{6}(2a-3t\pm{k})$

$Y=\frac{1}{6}(2a+3t\pm{k})$

$Z=\frac{1}{3}(a\mp{k})$

Thought the solution is determined by the equation Pell, but when calculating the sign was a mistake. There's no difference, but the amount should be. Therefore, the number of solutions of course.

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