I found this question in a video of the Cheenta Channel which I will link here for reference. It has been marked as a Simple yet Challenging problem in the end.
Here are my initial thoughts on the matter:
Let $P(x)=ax^3+bx^2+cx+d$ where $\overline {abcd}=p$ is a prime number.
Therefore $P(10)=p$
We can readily see that $a,b,c,d \in \{0,1,2,\dots,9\}$ where $a\neq 0$.
I tried to do it using the contrapositive assumption.I assumed that the polynomial has a rational root suppose $\frac{m}{n}$ where $m,n\in \Bbb Z$ and $n\neq 0$ and it is in lowest form, so
$P(x)=(nx-m)Q(x)$ where $Q(x)$ is a polynomial of degree $2$ which is true for all real $x$. On putting $x=10$ we see that
$P(10)=p=(10n-m)Q(10)$
And since $m$ and $n$ are integers which imply $(10n-m)$ is also an integer. And it must be $1$ since a $P(10)$ is a prime number. According to Rational Root Theorem which states:
Each rational solution $x = \frac{p}{q}$, written in lowest terms so that $p$ and $q$ are relatively prime, satisfies:
$p$ is an integer factor of the constant term, and
$q$ is an integer factor of the leading coefficient.
Therefore, $\left.m\right\rvert d$ and $\left.n\right\rvert a$ and the only combination working to make $(10n-m)=1$ is $m=9,n=1$ which tells us that the root is $9$. But it can not be possible since this cubic equation has positive integral coefficients and thus any positive root is not a possibility. We reach a contradiction.
Is there any flaw in this solution? If so please enlighten me. Thank you.
Best Answer
I think you have taken the main step. Let's assume the polynomial has a rational root. So, the polynomial is reducible over the rational numbers. Since $p$ is a prime number and no prime number divides all the coefficients $a,b,c,$ and $d$, by this lemma, we may assume: $$ax^3+bx^2+cx+d=(mx+n)(ex^2+fx+g),$$ where $m,n,e,f,$ and $g$ are integers. WLOG, we can assume $m,n>0$ because $ax^3+bx^2+cx+d$ cannot have any positive roots (note $\frac{-n}{m}$ is a root).
Now, on the one hand, we must have: $$9\geq a=me \implies 9\geq m\\ 9\geq d=gn \implies 9\geq n,$$
and on the other hand:
$$p=1000a+100b+10c+d=(10m+n)(100e+10f+g),$$ which is a contradiction because $p>1000>99\geq 10m+n>1$ while $p$ is a prime number. Therefore, the polynomial has no rational roots.