Prove that $\sqrt[n]{\frac{b^{n}}{a^{n}}+1}$ is irrational for every $b,a,n\in\mathbb{ N}$ and $n>2$. Prove that $\sqrt[n]{\frac{b^{n}}{a^{n}}+1}$ is irrational for every $b,a,n\in\mathbb{ N}$ and $n>2$.
I tried a long time using proof by contradiction and induction but it didn't work, it seem so difficult. I will be very thankful if someone helps me.
 A: We may assume without loss that $a$ and $b$ are coprime.
Suppose that $\rho=\sqrt[n]{\frac{b^{n}}{a^{n}}+1}$ is rational,
say $\rho=\frac{p}{q}$ where $p$ and $q$ are positive coprime
integers. Then
$$
\big(\frac{p}{q}\big)^n=\big(\frac{a}{b}\big)^n+1 \tag{1}
$$
So
$$
(pb)^n=(aq)^n+(qb)^n \tag{2}
$$
We see that $q^n$ divides $(pb)^n$, so $q$ divides $pb$, so
$q$ divides $b$ (since $p\wedge q=1$). Write $b=qc$ where $c$ is 
a positive integer. Then it follows from (2) that
$$
(pc)^n=a^n+(qc)^n \tag{3}
$$
Thus $c^n$ divides $a^n$, and hence $c$ divides $a$. So we see that
$c$ divides both $a$ and $b$ ; so $c=1$, and hence $b=q$. Then (2) becomes
$$
p^n=a^n+b^n
$$
We can then apply Fermat’s last theorem.
A: Proof could go like this. Let's suppose that $2$ integers exist $p,q$ (co-primes) such that
$$
\sqrt[n]{\frac{b^{n}}{a^{n}}+1}=\frac{p}{q}
$$
Then is easy to see that from the previous equation you get
$$
\frac{b}{a}=\sqrt[n]{p^n-q^n}/q
$$
so your question reduces to showing that, given two integers $p,q$ the number
$$
\sqrt[n]{p^n-q^n}
$$
is irrational or that two integers $m,l$ exist such that
$$
\sqrt[n]{p^n-q^n}=\frac{m}{l}
$$
or in other words
$$
(pl)^n=(ql)^n+m^n
$$
and we get a contradiction from the last Fermat theorem (if I have not done any mistake ;-) ) and therefore the starting expression cannot be rational.
A: $\sqrt[n]{\frac{b^{n}}{a^{n}}+1}=\sqrt[n]{\frac{b^{n}+a^n}{a^{n}}}=\frac{\sqrt[n]{b^{n}+a^n}}{a}$ But we know $\sqrt[n]{b^{n}+a^n}$ cannot be a rational number for $n>2$ according to Fermat's Last Theorem. Q.E.D
P.S: I doubt this would be solved without using Fermat's Last Theorem anyway.
