Prove that $f(x)=0$ has no rational solutions $f(x)$ $\in$ $Z[X]$ monic polynomial of degree $n$
$k,p$ $\in$ $N$
If none of the numbers $f(k), f(k+1), \ldots , f(k+p)$ is disivible by $p+1$, then $f(x)=0$ has no rational solutions.
 A: Hint $\ $ Suppose not, so $\,f(x)\,$ has a rational root $\,r.\,$ By the Rational Root Test, a rational root of a $\rm{\color{#c00}{monic}}$ polynomial $\in\Bbb Z[x]\,$ must be an integer, since the test implies that the denominator of a reduced rational root divides the leading coefficient $\color{#c00}{(= 1)}.\,$ Since  the $\,m = p+1\,$ consecutive integers $\,k,k\!+\!1,\ldots,k\!+\!p\,$ form a complete system of residues mod $\,m,\,$ the integer root $\,r\,$ must be congruent to one them, say $\, r \equiv k\!+\!i\pmod m.\, $ Therefore 
$\quad{\rm mod}\ m\!:\,\ k\!+\!i\equiv r\ \Rightarrow\ f(k\!+\!i)\equiv f(r) \equiv 0,\ $  i.e. $\ p\!+\!1 = m\mid f(k\!+\!i),\, $ contra hypothesis.
A: Hint: By the rational root theorem, all rational roots are integers.
Hint: $f(n+m) \equiv f(n) \pmod{m}$
A: Assume by contradiction that $f$ has a rational solution $a$. Then since $f$ is monic, the solution is integer.
Now, one of the numbers $k-a, k+1-a,.., k+p-a$ is divisible by $p+1$. Hence
$$p+1| (k+i)-a | P(k+i)-P(a)=P(k+i)$$
contradiction.
