An elliptic curve for $x_1^5+x_2^5+x_3^5=y_1^5+2y_2^5$?
问题内容
In a prior post, the equation,
$$x_1^5+2x_2^5 = y_1^5+2y_2^5$$
was considered. It has only one known primitive solution. This present post considers the similar,
$$x_1^k+x_2^k+x_3^k = y_1^k+2y_2^k$$
valid for both $k = (1,5)$. Duncan Moore found only one primitive solution, namely,
$$85333^k + 72631^k + 36981^k = 89703^k + 2\times52621^k$$
We can encode this in the form,
$$(a + c)^k + (-a + c)^k + (b + d)^k = (-b + d)^k + 2(b + c)^k $$
already true for $k=1$. Expanding for $k=5$, one must solve the quartic in $d$,
$$b (b^2 + d^2)^2 + c (a^2 + c^2)^2 = (b + c )(b^2 + c^2)^2$$
where,
$$d = \frac1b\sqrt{-b^4\pm b\sqrt{\color{blue}{b^6 - (a^4 - b^4) b c + 2 b^4 c^2 - 2 (a^2 - b^2) bc^3 + b^2 c^4}\,}}$$
The only rational solution $(a,b,c,d)$ so far is Moore's,
\begin{align} a &=+87\times73\\ b &=-87\times101\times3\\ c &=782\times101\\ d &=782\times81\end{align}
where the shared factors I think are not coincidence. The quartic polynomial in $c$ must be made a square,
$$b^6 - (a^4 - b^4) b c + 2 b^4 c^2 - 2 (a^2 - b^2) bc^3 + b^2 c^4 = y^2$$
With one initial rational point, it can be treated as an elliptic curve. I've found four polynomial solutions for $c$,
$$-b,\quad -\frac{(a^2+b^2)^2+4b^4}{4b(a^2+b^2)},\quad \frac{(a^2+b^2)^2}{4b^3},\quad -\frac{(a^2+b^2)(a^2-3b^2)}{4a^2b}$$
This causes the quartic in $d$ to factor into two quadratics, but seem unsolvable in the rationals. For example, let,
$$c = -\frac{(a^2+b^2)(a^2-3b^2)}{4a^2b}$$
and the quartic in $d$ has one factor solvable as,
$$d = \frac{(a^2+b^2)\sqrt{3(a^2+b^2)(a^2-3b^2)}}{4a^2b}$$
but there is no non-zero rational $(a,b)$ such that $d$ is also rational.
Question: Is there a rational polynomial $c=\dfrac{P_1(a,b)}{P_2(a,b)}$ of small degree less than $4$ such that the quartic in $d$ factors into two quadratics, one of which for the right $(a,b)$ is solvable in the rationals?
回答 (2)
Using the substitution, the fifth-power equation reduces to
$$b(b^2+d^2)^2+c(a^2+c^2)^2=(b+c)(b^2+c^2)^2.$$
This is the main equation to study.
Assume $b\neq 0$, and put
$$r=\frac{a}{b}, \qquad u=\frac{c}{b}, \qquad v=\frac{d}{b}.$$
After dividing by $b^5$, the equation becomes
$$(1+v^2)^2+u(r^2+u^2)^2=(1+u)(1+u^2)^2.$$
Therefore
$$(1+v^2)^2 = u^4-2(r^2-1)u^3+2u^2-(r^4-1)u+1.$$
Define
$$q(r,u)=u^4-2(r^2-1)u^3+2u^2-(r^4-1)u+1.$$
Then the condition is
$$q(r,u)=(1+v^2)^2.$$
This is the important point. It is not enough to make $q(r,u)$ a square. We need its square root to have the special form
$$1+v^2.$$
So the problem has two layers. First, we may study
$$z^2=q(r,u).$$
This is an elliptic curve in disguise. But the original problem also needs
$$z=1+v^2.$$
This second condition is essential.
If we take
$$u=-1.$$
This means
$$c=-b.$$
Substituting $u=-1$ into $q(r,u)$, we get
$$q(r,-1)=(r^2+1)^2.$$
Hence we may take
$$1+v^2=r^2+1.$$
So
$$v^2=r^2,$$
and therefore
$$v=\pm r.$$
Since $v=d/b$ and $r=a/b$, this gives
$$d=\pm a.$$
Therefore we have the family
$$\boxed{c=-b,\qquad d=\pm a.}$$
This works for every rational choice of $a$ and $b$, with $b\neq 0$.
Although the family above is valid, it is not a genuinely new solution. Indeed, if
$$c=-b,$$
then
$$y_2=b+c=0.$$
For example, if $d=a$, then
$$x_1=a-b, \qquad x_2=-a-b, \qquad x_3=a+b,$$
$$y_1=a-b, \qquad y_2=0.$$
The equation becomes
$$(a-b)^5+(-a-b)^5+(a+b)^5=(a-b)^5.$$
The two terms
$$(-a-b)^5 \qquad\text{and}\qquad (a+b)^5$$
cancel each other. Thus the family is correct, but it is trivial in spirit.
A non-trivial example
A much more interesting solution is obtained from
$$a=6351, \qquad b=-26361, \qquad c=78982, \qquad d=63342.$$
Then
$$x_1=a+c=85333,$$
$$x_2=-a+c=72631,$$
$$x_3=b+d=36981,$$
$$y_1=-b+d=89703,$$
$$y_2=b+c=52621.$$
Therefore
$$85333^5+72631^5+36981^5=89703^5+2\cdot 52621^5.$$
Also,
$$85333+72631+36981=89703+2\cdot 52621.$$
In normalized form, this solution is
$$r=\frac{a}{b}=-\frac{73}{303}, \qquad u=\frac{c}{b}=-\frac{782}{261}, \qquad v=\frac{d}{b}=-\frac{7038}{2929}.$$
For these values,
$$q(r,u)=(1+v^2)^2.$$
So this example satisfies both requirements: the elliptic-curve square condition and the extra condition needed to recover $d$.
Here the elliptic curve appears from
$$z^2=q(r,u).$$
But this is only the first part of the problem. The original fifth-power equation requires the stronger condition
$$z=1+v^2.$$
So the correct equation to study is
$$u^4-2(r^2-1)u^3+2u^2-(r^4-1)u+1=(1+v^2)^2.$$
The simple family
$$\boxed{c=-b, \qquad d=\pm a}$$
gives rational solutions of low degree, but these solutions are degenerate because $y_2=0$. For non-trivial solutions, one should search for rational points on
$$q(r,u)=(1+v^2)^2.$$
The example
$$(85333,72631,36981;89703,52621)$$
is a non-trivial solution of this type.
Hope this is helpful!
@Ataulfo
You are right that a quartic model
$$y^2=f(x)$$
gives an elliptic curve only when $f(x)$ is square-free, and one should check this condition. In the present case, the quartic is
$$q_r(u)=u^4-2(r^2-1)u^3+2u^2-(r^4-1)u+1.$$
Its discriminant is
$$-(r-1)^4(r+1)^4\left(32r^{10}+59r^8-28r^6-110r^4-348r^2-117\right).$$
Hence, for rational $r$, the quartic is square-free for
$$r\neq \pm 1.$$
Also, the curve has the rational point
$$(u,z)=(-1,r^2+1).$$
Therefore, for rational $r\neq \pm 1$, it is indeed birationally equivalent to an elliptic curve. The cases
$$r=\pm 1$$
are degenerate and should be excluded.
@David
Thank you. That numerical identity is interesting, but it belongs to a related, slightly different equation.
The identity should be written with the correct sign as
$$1355^5+2(526)^5=1685^5+2(-1349)^5.$$
Equivalently,
$$526^5+1355^5+526^5=1685^5-2(1349)^5.$$
So it gives a solution of the equation
$$a^5+2b^5=c^5+2d^5.$$
However, the present problem requires more than a fifth-power identity. It also requires the corresponding first-power identity. In other words, the same variables must work for both powers
$$k=1 \quad \text{and} \quad k=5.$$
For the suggested numerical identity, we would have
$$x_1=526, \qquad x_2=1355, \qquad x_3=526,$$
$$y_1=1685, \qquad y_2=-1349.$$
The fifth-power equation is true:
$$526^5+1355^5+526^5=1685^5+2(-1349)^5.$$
But the first-power equation is not true, because
$$526+1355+526=2407,$$
whereas
$$1685+2(-1349)=-1013.$$
Thus
$$2407\neq -1013.$$
This is why the example cannot be used directly in the parametrization
$$x_1=a+c, \qquad x_2=-a+c, \qquad x_3=b+d,$$
$$y_1=-b+d, \qquad y_2=b+c.$$
This parametrization automatically makes the first-power equation true.
Indeed, if one tries to force the first four values by setting
$$x_1=526, \qquad x_2=1355, \qquad x_3=526, \qquad y_1=1685,$$
then the parametrization gives
$$a=\frac{x_1-x_2}{2}=-\frac{829}{2},$$
$$c=\frac{x_1+x_2}{2}=\frac{1881}{2},$$
$$b=\frac{x_3-y_1}{2}=-\frac{1159}{2},$$
$$d=\frac{x_3+y_1}{2}=\frac{2211}{2}.$$
Therefore
$$y_2=b+c=-\frac{1159}{2}+\frac{1881}{2}=361.$$
So the parametrization would force
$$y_2=361,$$
not
$$y_2=-1349.$$
Hence the smaller numerical identity is useful for the different equation
$$a^5+2b^5=c^5+2d^5,$$
but it does not directly help with the simultaneous problem for
$$k=1,5.$$
For the present elliptic-curve construction, Duncan Moore's larger example remains the relevant one because it satisfies both powers.