Polynomial with no three distinct rational roots Prove that $x^3-2x^2-2x+a$, where $ a\in \mathbb{R} $, has no three distinct rational roots.
 A: Suppose 
$$f(x) = x^3 - 2x^2 - 2x + a = (x - P)(x - Q)(x - R)$$
where $P,Q,R$ are rational numbers. 

Our goal is to derive a contradiction.

If any of $P,Q,R$ is zero, then $a = 0$, hence $f(x) = x(x^2 - 2x - 2)$, contradiction, since $x^2 - 2x - 2$ has no rational roots.

Thus, assume $P,Q,R$ are nonzero.

By Vieta's formulas,
\begin{align*}
P + Q + R &= 2\\[6pt]
PQ + QR + RP &= -2
\end{align*}
Let $d$ be the least common denominator of $P,Q,R$. Then $d$ is a positive integer and
\begin{align*}
P &= \frac{p}{d}\\[6pt]
Q &= \frac{q}{d}\\[6pt]
R & = \frac{r}{d}
\end{align*}
where $p,q,r$ are nonzero integers such that $\text{gcd}(p,q,r) = 1$.

Then Vieta's formulas yield
\begin{align*}
p +  q + r &= 2d\\[6pt]
pq + qr + rp &= -2d^2\\[6pt]
\end{align*}
Then 
\begin{align*}
p^2 + q^2 + r^2 &= (p + q + r)^2 - 2(pq + qr + rp)\\[6pt]
&=(2d)^2 - 2(-2d^2)\\[6pt]
&=8d^2
\end{align*}
Since $p + q + r$  is even, and $\text{gcd}(p,q,r) = 1$, it follows that exactly one of $p,q,r$ is even.

Without loss of generality, assume $p$ is even, and $q,r$ are odd.

Recall that odd squares are congruent to $1$ mod $8$, and even squares are congruent to
$0$ or $4$ mod $8$. Then

\begin{align*}
&p^2 + q^2 + r^2 = 8d^2\\[6pt]
\implies\; &p^2 + q^2 + r^2 \equiv 0 \pmod 8\\[6pt]
\implies\; &p^2 + 1 + 1 \equiv 0 \pmod 8\\[6pt]
\implies\; &p^2 \equiv 6 \pmod 8
\end{align*}
contradiction. 

Therefore there do not exist rational numbers $P,Q,R$ such that
$$f(x) = x^3 - 2x^2 - 2x + a = (x - P)(x - Q)(x - R)$$
