# Determining the smallest possible value of a in a quadratic.

#### Ilikebugs

Numbers a, b and c form an arithmetic sequence if b − a = c − b. Let a, b, c be positive
integers forming an arithmetic sequence with a < b < c. Let f(x) = ax^2 + bx + c. Two distinct
real numbers r and s satisfy f(r) = s and f(s) = r. If rs = 2017, determine the smallest possible
value of a.

I tried changing b and c into a+d and a+2d to get (r-s)((r+s+1)a+(d+1))=0 but I am stuck on how to show the least possible value of a.

#### Idea

we need two equations

$$f(r)-f(s)=s-r$$

and

$$f(r)+f(s)=s+r$$

As you mentioned the first equation gives

$$1+b+a r+a s=0$$

or letting $t=r+s$,

$$1+b+a t=0$$

The second one can also be written in terms of $t$

$$a (t^2 - 4034) + b t - t + 2 c=0$$

Now eliminate $t$ in these two equations and substitute $b=a+d$ and $c=a+2d$ to get

$$d=\frac{-1-a+2016 a^2}{1+2 a}$$

Since $d$ is an integer, there are finitely many values possible for $a$

{9,26,503}

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