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Prove that the equation $x^3 -24x + k=0$ has one integer solution at most, $\forall k \in \mathbb{Z}$

Suppose there are two integer solutions. Then, according to Vieta, all the $3$ solutions are integers. Using the first derivative I can get a contradiction out of this assumption, but I'm only allowed to use elementary number theory, so I'm stuck here.

Any help is appreciated.

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  • $\begingroup$ "I'm only allowed to use elementary calculus": are you missing a word here? $\endgroup$ – Erick Wong Aug 1 '17 at 17:32
  • $\begingroup$ @ErickWong I don't know, just tell me that do you think is missing. $\endgroup$ – user261263 Aug 1 '17 at 17:55
  • $\begingroup$ If you were only allowed to use elementary calculus, then there would be no problem using the first derivative. Did you mean to say you're only allowed elementary algebra? Elementary number theory? Or not allowed to use calculus? $\endgroup$ – Erick Wong Aug 1 '17 at 17:57
  • $\begingroup$ @ErickWong Elementary number theory is what I meant actually. $\endgroup$ – user261263 Aug 1 '17 at 17:59
  • $\begingroup$ Thanks, I have edited the question to fix this. Though I am curious how it could be done with calculus, given that the claim doesn't hold for many numbers different from $24$. Hard to see how calculus can resolve that distinction. $\endgroup$ – Erick Wong Aug 2 '17 at 16:58
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If there are three integer solutions, they add to $0$ and so are $a$, $b$ and $c=-a-b$. Then $$-24=ab+ac+bc=ab-(a+b)^2=-a^2-ab-b^2.$$ You need to prove that $24$ is not represented over the integers by the quadratic form $a^2+ab+b^2$. Maybe you could complete the square...

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  • $\begingroup$ $$96=4*24=4a^2+4ab+4b^2=(2a+b)^2+3b^2$$ Now $3|2a+b$ and hence $$32=3x^2+y^2$$ Looking modulo $8$ you can deduce that both $x,y$ are even. Divide by 4 and repeat: look modulo 8, both number are even.... $\endgroup$ – N. S. Aug 1 '17 at 15:55
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    $\begingroup$ $a^2+ab+b^2=24$ has no solutions mod $9$. $\endgroup$ – lhf Aug 1 '17 at 16:14
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It is simpler: Notice $x(x^2-24)=k$ is a unique representation of $k$ since function $x$ is injective for any $x$ and $(x^2-24)$ is injective for positive $x$. (Prime factorisation theorem).

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  • $\begingroup$ This argument is unsound: the same properties hold for the factors of $x(x^2-7)$, but the equation $x(x^2-7) = -6$ has three integer solutions ($x=-3,-2,1$). $\endgroup$ – Erick Wong Aug 1 '17 at 17:21
  • $\begingroup$ Missed it, you're right. $\endgroup$ – Toni Mhax Aug 1 '17 at 17:33

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