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Find all integer solutions $(x,y)$ of the equation $$400x^6-60x^3-343y^3+3 = 0.$$

One solution is $(x,y) = (1,1)$. Also, $343 = 7^3$, so we can write $$400x^6-60x^3+3 = (7y)^3.$$ How can we prove that $(x,y) = (1,1)$ is the only solution?

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    $\begingroup$ Equations of such a high degree can be very hard to solve. Let $a=20x^3$, $b=7y$, so we get $a^2-3a+3=b^3$. That might be easier to deal with. Maybe you can show the only solution is $a=20$, $b=7$. $\endgroup$ – Gerry Myerson May 21 '17 at 0:49
  • $\begingroup$ Where do your problem comes from ? $\endgroup$ – Jean Marie May 21 '17 at 3:56
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    $\begingroup$ The simplified equation a^2-3a+3=b^3 with the solution (20,7) also has the solutions (1,1), (2,1), (-17,7). There may be more, but I don't whether any of them could give integer points on the original curve. Note that it is an elliptic curve, so It certainly has an infinite number of rational solutions, because they form a group. $\endgroup$ – Jaap Scherphuis May 21 '17 at 4:35
  • $\begingroup$ An infinite number of rational solutions, but only finitely many integer solutions. $\endgroup$ – Gerry Myerson May 21 '17 at 12:12
  • $\begingroup$ I solve your question. so I edit my answer. $\endgroup$ – G.H.lee May 23 '17 at 4:26
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Let $x^{3}= X , 7y = Y.$ Then

$400X^{2} - 60X + 3 - Y^{3} = 0 $

$X = \frac{3\pm \sqrt{4Y^{3}-3}}{40}$

and $Z^{2} = 1372y^{3} - 3$ then $x^{3} = \frac{3\pm Z}{40}$

x is integer so $ Z\pm 3 = 40 , 320, 1080 , ... \Rightarrow 1372y^{3} - 3 = 37^{2} , 43^{2} ,317^{2}, 323^{2}, ...$

and set $ S = \left \{37 , 43, 317, 323 , 1077, 1083 ... \right \}$ , $ y = 1 + y' $

then $1372y'^{3} + 4116 y'^{2} + 4116y' + 1369 = 37^{2} + (s^{2} - 37^{2})$

$\Rightarrow 1372y' \cdot (y'^{2} + 3 y' + 3) = (s^{2} - 37^{2})$ when $s \in S $

and $1372 = 7^{3} \cdot 2^{2} $ , $s^{2} - 37^{2} = (s+37)(s-37)$

$ s \in S \Rightarrow s $ is an odd number. so $4 | s^{2} - 37^{2}$

$\frac{(s^{2} - 37^{2}) }{ 4} = s'$ then

$ 343y' \cdot (y'^{2} + 3 y' + 3) = s'$

if $ y' \neq 0$ then $ 343|s'$ but $ s= 40 \cdot x^{3} \pm 3$

I don't know how to solve your problem, but it may be helpful.

If your equation has other solutions, then $ 7^{3}|(s^{2} - 37^{2})$ (When $ s= 40 \cdot x^{3} \pm 3$ , )

(I think that s does not exist)

p.s. I'm sorry that I'm poor at English.


I prove $ 343\mid s^2 - 37^2 $ when $ s = 40x^3 \pm 3 $

First: $ s = 40x^3 - 3 $

$ s^2 - 37^2 = 80(x^3 - 1)(20x^3 + 17) $

If $ 7\mid x^3 - 1 $ , $ 7\nmid 20x^3 + 17 $

because $ 20x^3 + 17 = 20(x^3 - 3 ) + 77$, but $ 7\nmid x^3 - 3 (\because 7\mid x^3 - 1 )$

so , $ 343\mid s^2 - 37^2 $ need $ 343\mid 20x^3 + 17 $

but $ x^3 \mod 7 $ can be only 0,1,6

so $ 7\nmid 20x^3 + 17 $

Finally, we proved $ 343\nmid s^2 - 37^2 $ when $ s = 40x^3 - 3 $


Second: $ s = 40x^3 + 3 $

$ s^2 - 37^2 = 80(x^3 + 1)(20x^3 - 17) $

If $ 7\mid x^3 + 1 $ , $ 7\nmid 20x^3 - 17 $

because $ 20x^3 - 17 = 20(x^3 + 3 ) - 77$, but $ 7\nmid x^3 + 3 (\because 7\mid x^3 + 1 )$

so , $ 343\mid s^2 - 37^2 $ need $ 343\mid 20x^3 - 17 $

but $ x^3 \mod 7 $ can be only 0,1,6

so $ 7\nmid 20x^3 + 17 $

Finally, we proved $ 343\nmid s^2 - 37^2 $ when $ s = 40x^3 \pm 3 $

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  • $\begingroup$ $Z^2=4Y^3-3$, $(4Z)^2=(4Y)^3-3$ is a Mordell equation, there are tabulations of integer solutions on the web. $\endgroup$ – Gerry Myerson May 21 '17 at 12:14

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