Putnam 2011, Problem B-2

Let $S$ be the set of all ordered triples $(p,q,r)$ of prime numbers for which at least one rational number $x$ satisfies $px^2 + qx + r = 0$. Which primes appear in seven or more elements of $S$?

Solution:
Only the primes \(2\) and \(5\) appear seven or more times. The fact that these primes appear is demonstrated by the examples  
\[
(2,5,2), (2,5,3), (2,7,5), (2,11,5)
\]  
and their reversals.  
It remains to show that if either \(\ell = 3\) or \(\ell\) is a prime greater than \(5\), then \(\ell\) occurs at most six times as an element of a triple in \(S\). Note that \((p,q,r) \in S\) if and only if \(q^2 - 4pr = a^2\) for some integer \(a\); in particular, since \(4pr \geq 16\), this forces \(q \geq 5\). In particular, \(q\) is odd, as then is \(a\), and so \(q^2 \equiv a^2 \equiv 1 \pmod{8}\); consequently, one of \(p,r\) must equal \(2\). If \(r = 2\), then \(8p = q^2 - a^2 = (q + a)(q - a)\); since both factors are of the same sign and their sum is the positive number \(2q\), both factors are positive. Since they are also both even, we have \(q + a \in \{2,4,2p,4p\}\) and so \(q \in \{2p + 1, p + 2\}\). Similarly, if \(p = 2\), then \(q \in \{2r + 1, r + 2\}\). Consequently, \(\ell\) occurs at most twice as many times as there are prime numbers in the list  
\[
2\ell + 1, \ell + 2, \frac{\ell - 1}{2}, \ell - 2.
\]  
For \(\ell = 3\), \(\ell - 2 = 1\) is not prime. For \(\ell \geq 7\), the numbers \(\ell - 2, \ell, \ell + 2\) cannot all be prime, since one of them is always a nontrivial multiple of \(3\).  
**Remark.** The above argument shows that the cases listed for \(5\) are the only ones that can occur. By contrast, there are infinitely many cases where \(2\) occurs if either the twin prime conjecture holds or there are infinitely many Sophie Germain primes (both of which are expected to be true).