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I know it's probably easy, but don't know how to prove exactly. I know roots are the divisors of a0/an and about viete, but I don't know if the proof is enough, please help

Not important
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  • But $a_n = 1$. That's a huge limitation. And $a$ is odd.... so what about $x^3 - x +a=(x+1)x(x-1) + a$. Can that ever equal $0$ if $x|a$ and $a$ is odd. What about $x$? Is $x$ an integer? is it odd or even? Does it matter? – fleablood Apr 20 '20 at 17:17
  • By the rational root test any rational root is an integer, but there are no integer roots by the Parity Root Test in the linked dupe. – Bill Dubuque Apr 21 '20 at 02:47

3 Answers3

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Any rational root would have to be an odd integer. But if $x$ is an odd integer, $x^3 - x + a$ is odd.

Robert Israel
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By the rational root theorem the roots would have to be odd integers (because they would be divisors of $a$), and odd integers give an odd result.

Asinomás
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  • so if we get like 2 roots with the rational root theorem, then the third one can't be rational, it can be complex or something else? – Not important Apr 20 '20 at 17:04
  • Don't think about factoring the polynomial, just think about the fact that $Q(x)$ cannot be $0$ if $x$ is an odd integer. – Asinomás Apr 20 '20 at 17:06
  • yes, that is true, but you don't get all roots with the theorem, why can't the remaining ones be rational? sorry if it's dumb from me – Not important Apr 20 '20 at 17:11
  • Well, your original problem asks about rational roots, and the theorem tells you that rational roots are odd integers. We do not care about non-rational roots for the purpose of your problem. – Asinomás Apr 20 '20 at 17:12
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    oh ok, I understood the point , thanks a lot – Not important Apr 20 '20 at 17:14
  • "why can't the remaining ones be rational" The only possible denominators are $1$ (and negative $1$). – fleablood Apr 20 '20 at 17:22
  • "then the third one can't be rational, it can be complex or something else?" Or irrational real. Actually you can't have just one complex root (but that's another story). And you don't get 2 rational roots. You get $0$ rational roots. Either there are $3$ irrational real roots, or $1$ irrational real root and $2$ complex roots. But we weren't asked about that. We were only asked to show there there are no rational roots; we weren't asked anything about what the roots actually are. – fleablood Apr 20 '20 at 17:26
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It's sortof not really a trick question.

The rational root theorem so that if there is a rational root it is of the form $\pm \frac mn$ where $m|a$ and $b|1$.

The only thing that divides $1$ is $\pm 1$ so if there is a rational root it is $\pm m$ where $m|a$.

So if there is any rational root it is an integer.

What's more, as $a$ is odd then $m|a$ means $m$ is odd and so if there is any rational root it is an odd integer.

.....

But the thing is: If $x$ is an odd integer then $x^3 - x$ will always be even.

And if $a$ is not even then $x^3 - x + a$ will not be even. So $x^3 - x +a =0$ is impossible.

Recap:

By rational root theorem the only rational roots of polynomial with leading coefficent of $1$ will be integers.

For any integer input, then $x^3 - x + a$ will always be odd and thus can not have any integer roots.

fleablood
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