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

Question

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?

Answer 1

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.

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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 $

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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 $