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:
- The first and the second parts of the sum are equal to 0.
- 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.
Best Answer
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 $n=1$.