Mathematical Reflections 2016, Issue 6 - Problem S395

Problem:
Let $a,b,c$ be positive integers such that $$a^2b^2+b^2c^2+c^2a^2-69abc=2016.$$
Find the least possible value of $\min(a,b,c)$.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Assume without loss of generality that $a \leq b \leq c$. If $a=1$, then $$b^2+b^2c^2+c^2-69bc=2016.$$ This equation is equivalent to
$$(2bc-67)^2+4(c-b)^2=12553.$$ It follows that $0 \leq c-b \leq 56$ and $0 < bc \leq 89$. Hence $b^2 \leq 89$, which gives $b \leq 9$. An easy check gives no positive integer solutions for $c$. If $a=2$, then $$4b^2+b^2c^2+4c^2-138bc=2016.$$
This equation is equivalent to $$(bc-65)^2+4(c-b)^2=6241.$$ It follows that $0 \leq c-b \leq 39$ and $0 < bc \leq 144$. Hence $b^2 \leq 144$, which gives $b \leq 12$. An easy check gives $b=c=12$, so the least possible value of $\min(a,b,c)$ is $2$.