M5. Consider equations of the form $a x^{2}+b x+c=0$, 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?
Solution
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 $(\ldots)(\ldots)$.
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 $a x$ and $x$. Hence the factorisation is either $(a x+c)(x+1)$, in which case $b=a+c$, or $(a x+1)(x+c)$, in which case $b=a c+1$. 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 $b=a c+1$. 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.