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Question:

Show that $30x^n-91$ does not have rational roots for $n\in \mathbb{N}$ where $n>1$.

My Attempt:

Let's prove it by induction on n. For $n=2$ we have the following roots: $$ r=\frac{\pm \sqrt{10920}}{60}\notin \mathbb{Q} $$ Now suppose that it is true that it doesn't have any rational roots for $n<j$. For $n=j$ it follows that: $$ 30x^j-91 = x(30x^{j-1}-91)+91(x-1) $$ hence there are two possibilities for it to be equals to 0:

  1. The first and the second parts of the sum are equal to 0.
  2. The first part of the sum is the symmetric element of the second part

Now let's apply our induction hypothesis: $$ 30x^j-91 = x\underbrace{(30x^{j-1}-91)}_{\neq 0}+91(x-1) $$ For the first statement to be true, we'll need to have $x=0$ since $30x^{j-1}-91\neq 0$ for every $x \in \mathbb{Q}$. But $x=0$ results in $-91$ in the second part of the sum, therefore that possibility is discarded.

For the second statement to be true, we'll need to have: $$ x\underbrace{(30x^{j-1}-91)}_{\neq 0}=-91(x-1) $$ Let's suppose that is true, which means that for every value of $x\in \mathbb{Q}$ we'll need to have an equality.

Plugging in $x=0$ we get $0$ at the LHS and $91$ at RHS, therefore statement 2 is not true and we're done.

What do you think about what I've done? How'd you do it?

Thank you and any constructive critics and help are highly appreciated.

Bruno Reis
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  • It's not valid to conclude that $x(30x^{j-1}-91) = -91(x-1)$ has to be true for every value of $x \in \mathbb Q$. From the assumptions in that case you only know that there's some such $x$. – Misha Lavrov Nov 10 '18 at 00:36
  • Do you know the rational root test or Eisenstein? Or you could use unique factorization of rationals to deduce the form of nth powers, or proceed similarly as in irrationality proofs of square-roots, etc. – Bill Dubuque Nov 10 '18 at 00:36
  • I would just prove this directly by assuming $p/q$ solves the equation for integers $p,q$ with no common factors, and reach a contradiction. No need for induction. – Michael Nov 10 '18 at 00:36
  • @BillDubuque nope! – Bruno Reis Nov 10 '18 at 00:38
  • @MishaLavrov Yeah, you are right! How can I show, following the proof from there, that there isn't such $x \in \mathbb{Q}$? – Bruno Reis Nov 10 '18 at 00:39
  • Compare (unique) prime factorizations or, equivalently, use Euclid's Lemma, after clearing denominators. – Bill Dubuque Nov 10 '18 at 00:41
  • Why such a complicated approach when you know that the solutions are the irrationals $\sqrt[n]{91/30}$ ? –  Nov 10 '18 at 01:00
  • When you have a hammer everything looks like a nail. There is utterly no reason Induction should have seemed like a good idea. There is no natural intuition how having no solutions in $n$ would have anything to do with finding solutions in $n+1$ so induction is a terrible idea. Instead $30x^2 -91=0\implies x = \sqrt[n]{\frac {91}{30}}$ (as Yves Daoust suggests) is a much better idea. – fleablood Nov 10 '18 at 01:07
  • @fleablood Yeah, induction was unappropriated here... I'm just learning and haven't created those intuitions yet. Ty anyway – Bruno Reis Nov 10 '18 at 01:10
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    It was a valiant effort but... doomed to failure. – fleablood Nov 10 '18 at 01:20
  • @fleablood couldn't agree more! – Bruno Reis Nov 10 '18 at 01:25
  • @fleablood Actually induction can be used to prove things like this, e.g. see the proofs of the Rational Root Test I gave here and here – Bill Dubuque Nov 10 '18 at 01:38

2 Answers2

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Well if $\frac pq; \gcd (p,q) =1;p,q$ integers is a root then

$30\frac {p^n}{q^n} = 91$. As $91$ is an integer, and $p$ and $q$ are relatively prime, $q^n|30$.

If $q = 1$ then $30p^n = 91$ which has no integer solutions.

If $q \ne 1$ then $q^n|30=2*3*5$. But $30 = 2*3*5$ has no prime factors of powers greater than $1$. So $n= 1$. and $\frac pq = \frac {91}{30}$. That is is the only rational solution and it only holds for $n=1$.

Bill Dubuque
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fleablood
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  • Nice approach. Thanks! – Bruno Reis Nov 10 '18 at 01:03
  • @BrunoReis Be sure you understand the implicit use of Euclid's Lemma (or unique prime factoirization) in the above method. Your instructor may require you to be more explicit to be rigorous. – Bill Dubuque Nov 10 '18 at 01:04
  • @BillDubuque Yeah, I've got it! I was just trying a different approach... Thank you – Bruno Reis Nov 10 '18 at 01:06
  • Yes, Unique Prime factorization must utterly be assumed. (Or proven elsewhere). – fleablood Nov 10 '18 at 01:09
  • @fleablood we've proved it in class in different ways a while ago. There are so many elegant proofs... – Bruno Reis Nov 10 '18 at 01:12
  • This would be a pretty pointless problem to attempt to prove without knowing unique prime factorization (although it could be done). – fleablood Nov 10 '18 at 01:16
  • @fleablood Actually it can be done very simply without - see my comment on lhf's answer. Btw, typo: you mean p & q, not p & n are coprime (hence so too are p^n & q^n) – Bill Dubuque Nov 10 '18 at 01:21
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If $30x^n-91$ has a rational root, then $30a^n=91b^n$ for some $a,b \in \mathbb Z$.

Considering the maximal power of $2$ that divides each side, we get $1+n\alpha=n\beta$, which is a contradiction since $n>1$ does not divide $1$.

lhf
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    Remark that this method doesn't require full-blown uniqueness of prime factorizations, only factorizations of the form $,2^k n,$ for $n$ odd - which is much more elementary. – Bill Dubuque Nov 10 '18 at 00:55
  • Thank you for answering @lhf but for me, the other answer was more intuitive! Nice approach too! – Bruno Reis Nov 10 '18 at 01:04