Mathematical Reflections 2017, Issue 5 - Problem O422

Problem:
Let $P(x)$ be a polynomial with integer coefficients having an integer root $k$ and $P(0) \neq 0$. Prove that if $p$ and $q$ are distinct odd primes such that $P(p)=p<2q-1$ and $P(q)=q<2p-1$, then $p$ and $q$ are twin primes.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Assume that $k \in \mathbb{Z}$ is a root of the polynomial with integer coefficients $P$. So, $P(k)=0$, which gives $P(x)=(x-k)Q(x)$ for some polynomial with integer coefficients $Q(x)$. Hence,
$$p=P(p)=(p-k)Q(p), \qquad q=P(q)=(q-k)Q(q).$$ It follows that $(p-k) \mid p$ and $(q-k) \mid q$, so $p-k \in \{\pm 1, -p\}$ and $q-k \in \{\pm 1, -q\}$. Since $p$ and $q$ are distinct primes, then $p-k \neq q-k$ and we have seven cases.

(i) $p-k=-1$ and $q-k=1$. It follows that $q-p=2$, i.e. $q=p+2$, so $p$ and $q$ are twin primes.

(ii) $p-k=-1$ and $q-k=-q$. It follows that $p=2q-1$, contradiction.

(iii) $p-k=1$ and $q-k=-1$. It follows that $p-q=2$, i.e. $p=q+2$, so $p$ and $q$ are twin primes.

(iv) $p-k=1$ and $q-k=-q$. It follows that $p=2q+1$, contradiction.

(v) $p-k=-p$ and $q-k=-1$. It follows that $q=2p-1$, contradiction.

(vi) $p-k=-p$ and $q-k=1$. It follows that $q=2p+1$, contradiction.

(vii) $p-k=-p$ and $q-k=-q$. It follows that $p-q=-p+q$, i.e. $p=q$, contradiction.

In conclusion, $p$ and $q$ are twin primes.