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Find all $$x,y\in\mathbb{Z}$$ such that $$2x^3-7y^3=3$$

Solution:

We consider first $$2x^3-7y^3\equiv3 \pmod 2$$ $$5y^3\equiv 1 \pmod 2$$ $$y^3\equiv 1 \pmod2$$ which has solution $y\equiv 1 \pmod 2$

Consider $$2x^3\equiv 3 \pmod 7$$ $$4\cdot 2x^3\equiv 4\cdot3 \pmod 7$$ $$x^3\equiv 5 \pmod 7$$

but none of $x=0,1,2,3,4,5,6\pmod 7$ satisfies the equation

Which would mean that the equation has no integer solutions. Can this be correct?

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  • $\begingroup$ Related : math.stackexchange.com/questions/603104/… $\endgroup$ – lab bhattacharjee Jan 16 '14 at 18:16
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    $\begingroup$ Yes, this can be correct (and it is). Since the congruence modulo $7$ has no solution, there are no solutions. $\endgroup$ – Daniel Fischer Jan 16 '14 at 18:17
  • $\begingroup$ @DanielFischer: Ah that is nice:) Thanks. $\endgroup$ – H.E Jan 16 '14 at 18:19
  • $\begingroup$ @DanielFischer or J.W.S.: Please post an answer (and J.W.S. please accept it) so this falls off the "no answers" list. $\endgroup$ – Kieren MacMillan Sep 15 '14 at 17:21
  • $\begingroup$ @KierenMacMillan J.W.S. hasn't been seen since April, so I don't expect him/her to visit again soon. I converted the comment to an answer. $\endgroup$ – Daniel Fischer Sep 15 '14 at 17:33
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Indeed, the equation $2x^3 - 7y^3 = 3$ has no integer solutions. Modulo $7$, we get the congruence

$$2x^3 \equiv 3 \pmod{7}$$

which every solution of the original equation must satisfy. But the group of nonzero remainders modulo $7$ has order $6$, so $x^3 \equiv \pm 1 \pmod{7}$ for all $x\not\equiv 0 \pmod{7}$. Thus, whatever $x$ is, the remainder of $x^3$ modulo $7$ is one of $-1,0,1$, and so $2x^3$ has remainder $0,2$, or $5$, so $$2x^3\not\equiv 3 \pmod{7}$$

for all $x\in\mathbb{Z}$.

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