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.
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 $n=1$.