Maclaurin 2015 Q5
5. Consider equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), \(c\) are all single-digit prime numbers.
How many of these equations have at least one solution for \(x\) that is an integer?
Maclaurin Olympiad 2015, Question 5. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: 7
Note that \(a\), \(b\) and \(c\), being prime, are positive integers greater than 1.
If the equation has an integer solution, then the quadratic expression factorises into two ‘linear’ brackets (...)(...).
Since \(c\) is prime, the ‘constant’ terms in the brackets can only be \(c\) and 1. Similarly, since \(a\) is prime, the \(x\) terms can only be \(ax\) and \(x\). Hence the factorisation is either \((ax + c)\,(x + 1)\), in which case \(b = a + c\), or \((ax + 1)\,(x + c)\), in which case \(b = ac + 1\). Each case leads to an integer solution for \(x\), either \(x = -1\) or \(x = -c\).
In the first case exactly one of \(a\) or \(c\) is 2, since \(b\) is prime and \(b = a + c\). Therefore \((a, b, c)\) is one of (2, 5, 3), (2, 7, 5), (3, 5, 2) or (5, 7, 2).
In the second case one or both of \(a\) and \(c\) is 2, since \(b\) is prime and \(b = ac + 1\). Therefore \((a, b, c)\) is one of (2, 5, 2), (2, 7, 3) or (3, 7, 2).
Hence altogether there are seven such quadratic equations.
Solution © UK Mathematics Trust, from the Maclaurin Olympiad 2015 solutions.