The quadratic equation x^2 - ax + 2016 = 0 has two postiive integer solutions. Find the minimum value of a.
We let the positive integer solutions to be \(\alpha\) and \(\beta\). Without loss of generality, assume \(\alpha < \beta\).
Then,
\(x^2 - ax + 2016 = (x- \alpha)(x - \beta)\\ x^2 - ax + 2016 = x^2 - (\alpha + \beta)x + \alpha \beta\)
So, the question is actually asking for the minimum value of \(\alpha + \beta\) for \(\alpha \beta = 2016\). We can list the factors of 2016.
Factors of 2016:
\(\alpha\) | 1 | 2 | 3 | 4 | 6 | 7 | 8 | 9 | 12 | 14 | 16 | 18 | 21 | 24 | 28 | 32 | 36 | 42 |
\(\beta\) | 2016 | 1008 | 672 | 504 | 336 | 288 | 252 | 224 | 168 | 144 | 126 | 112 | 96 | 84 | 72 | 63 | 56 | 48 |
I have listed them in a unique way such that on each column, the numbers multiply to 2016.
We can see that a is the minimum sum of these pair of numbers, which is 42 + 48 = 90.