Vieta's Formula Only Plays A Small Role

Lemma 1

Suppose for a fixed $c$ that $a,b$ satisfies the equation. Then $b,\frac{b^{2}+2}{a}$ is also a solution

Proof 1

We know that $a^{2}-abc+b^{2}+2=0$ . Let $a=x$ . Then our polynomial becomes $x^{2}-bcx+b^{2}+2=0$ and has solutions $a,d$ for some value of $d$ .

We know by Vieta's Formula that $a+d=bc$ , and since $bc-a$ is an integer, $d$ must also be an integer.

Vieta's formula also tells us that $ad=b^{2}+2$ , so we know that $d=\frac{b^{2}+2}{a}$ . Hence, $b,\frac{b^{2}+2}{a}$ is also a valid solution $\square$ .

Lemma 2

If $a>b\geq3$ , then $0<d<b$ .

Proof 2

The fact that $d=\frac{b^{2}+2}{a}$ is a positive integer divided by a positive integer tells us that $d>0$ .

$d=\frac{b^{2}+2}{a}<\frac{b^{2}+2}{b}=b+\frac{2}{b}<b+1$ . If $d$ is an integer, and $b+1$ is an integer, then $d<b+1$ implies that $d\leq b$ .

If $d=b$ , then substituting this into the original equation gets that $2b^{2}+2-b^{2}c=0 \rightarrow 2=b^{2}(c-2)$ , and it is easy to see that our only solution will be $d=b=1, c=4$ , but $b\geq3$ , a contradiction.

Hence, $0<d<b$ $\square$ .

Now , from the above information we can conclude that if a solution exists, call it $A_{k},B_{k},C_{1}$ , such that $A>B\geq3$ , we can "reduce" this solution to another solution $A_{k-1},B_{k-1},C_{1}$ , and continue this process until we reach $A_{1},B_{1},C_{1}$ , where $b<3$ . Hence, we only need to check for possible values of $c$ at $b=1,2$ .

$b=1$

$a^{2}+1-ac+2=0 \rightarrow a^{2}+3=ac \rightarrow a+\frac{3}{a}=c$ . Since in order for $c$ to be an integer, $\frac{3}{a}$ must also be an integer, so we get that $a=1,3$ are possible solutions. Either case gives us $c=4$ .

$b=2$

$a^{2}+4-2ac+2=0 \rightarrow a^{2}+6=2ac \rightarrow a+\frac{6}{a}=2c$ . By the same logic as before we can conclude that $a=1,2,3,6$ are all possible values of $a$ , and we can then conclude that $2c=5,7$ . In either case, $c$ is not an integer, so we can reject this case.

Lastly , since we know that $c=4$ is the only possible value of $c$ , we can simply work our way back up a chain $A_{n}.B_{n},4$ until one of the values surpasses 2016. We do this by using the same relation of our decreasing sequence in reverse, being $b_{n+1}=\frac{a_{n}^{2}+2}{b_{n}}$ , and interchanging $a$ and $b$ as we go.

$\left(A_{1},B_{1}\right)=\left(1,1\right) \\ \left(A_{1}.B_{2}\right)=\left(\frac{1^{2}+2}{1},1\right) = \left(3,1\right) \\ \left(A_{2},B_{2}\right)=\left(3,\frac{3^{2}+2}{1}\right)=\left(3,11\right) \\ \left( A_{2},B_{3}\right)=\left( \frac{11^{2}+2}{3},11\right)=\left( 41,11\right) \\ \left(A_{3},B_{3}\right)=\left( 41,\frac{41^{2}+2}{11}\right)=\left( 41,153\right) \\ \left(A_{3},B_{4}\right)=\left(\frac{153^{2}+2}{41},153\right)=\left(571,153\right) \\ \left(A_{4},B_{4}\right)=\left(571,\frac{571^{2}+2}{153}\right)=\left(571,2131\right)$

Note that each of the $a=1$ solutions appear in this chain, so we can conclude that the other chain will give us the same results.

Since $2131>2016$ , our answer is $\boxed{571}$ .