Algebra: Rational roots of a polynomial of degree $4$ Consider the following polynomial with real coefficients: $x^4(t^5 - 1) - x^3(1+4t^5) + x^2(6t^5 -1) - x(1+4t^5) + (t^5 -1 ) = 0$, where $x$ and $t$ are both rational and $t$ is fixed.
By simple algebra, we find that the sum of the roots is $\dfrac{1+4t^5}{1-t^5}$ and their product is $1$. Prove that any rational solution $x$ to this equation is in fact an integer.
From my initial attempts, the ideas of the sum and product of the roots seem not to be offering much help.

Let $x$ be a rational solution to the equation. We can easily check that $x \not = 1$ and $t \not = 0$. Then
$$\begin{array}{c c c c}
&x^4 (t^5-1) - x^3 (4t^5+1) + x^2 (6t^5-1) - x (4t^5+1) + (t^5-1)&=&0 \\
\Leftrightarrow&t^5 (x^4 - 4x^3 + 6x^2 - 4x + 1) - (x^4 + x^3 + x^2 +x + 1) &=&0\\
\Leftrightarrow&t^5 (x-1)^4&=&x^4 + x^3 + x^2 +x + 1\\
\Leftrightarrow&t^5 (x-1)^5&=&(x-1)(x^4 + x^3 + x^2 +x + 1)\\
\Leftrightarrow&t^5 (x-1)^5&=&x^5-1
\end{array}$$
Now let's have $t = \dfrac{p}{q}$ and $x = \dfrac{a}{b}$. This gives us
$$\begin{array}{c c c c}
&\dfrac{p^5}{q^5} (\dfrac{a}{b}-1)^5&=&\dfrac{a^5}{b^5}-1\\
\Leftrightarrow&p^5 \dfrac{(a-b)^5}{b^5}&=&\dfrac{a^5-b^5}{b^5} q^5\\
\Leftrightarrow&p^5 (a-b)^5&=&(a^5-b^5) q^5\\
\end{array}$$
The last equality implies that there exists $c \in \mathbb{Z}$ such that $a^5-b^5=c^5$.
By Fermat's Last Theorem, we either have $a = 0$ or $b = 0$ or $c = 0$. The last two cases are impossible since $b \not = 0$ and $x \not = 1$. Hence we must have $a = 0$.
$$\boxed{\text{So the only possible rational solution to the equation is }x = 0}$$
And we can easily see it is indeed a solution if and only if $t = 1$.
