MHB What is the strategy for solving Problem S395 in Mathematical Reflections?

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The problem S395 from Mathematical Reflections involves the equation $a^2b^2+b^2c^2+c^2a^2-69abc=2016$, where $a, b, c$ are positive integers. Traditional inequalities like AM-GM and Cauchy-Schwarz have proven ineffective in finding a solution. A suggested approach is to consider minimizing the values of $a, b, c$ by testing small integers starting from 1. The deadline for submission of solutions is January 15, 2017, prompting caution in sharing insights. Exploring the least values systematically may yield potential solutions.
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If $a^2b^2+b^2c^2+c^2a^2-69abc=2016$, then, what can be said about the least value of $\min(a, b ,c)$?

This problem is unyielding to the major inequalities like AM-GM, Cauchy-Schwarz, etc. I also tried relating it to $x^3+y^3+z^3-3xyz=(x+y+z)(\sum_{cyc}x^2+\sum_{cyc}xy)$, but of no use. Any ideas. Thanks beforehand.

PS: This is problem S395 in Mathematical Reflections.
 
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vidyarth said:
If $a^2b^2+b^2c^2+c^2a^2-69abc=2016$, then, what can be said about the least value of $\min(a, b ,c)$?

This problem is unyielding to the major inequalities like AM-GM, Cauchy-Schwarz, etc. I also tried relating it to $x^3+y^3+z^3-3xyz=(x+y+z)(\sum_{cyc}x^2+\sum_{cyc}xy)$, but of no use. Any ideas. Thanks beforehand.

PS: This is problem S395 in Mathematical Reflections.
First, you need to include the information (given in the statement of Problem S395) that $a,b,c$ are positive integers. Without that information the problem does not make much sense.

Second, that section of Mathematical Reflections says that this problem is in a list whose deadline for submissions is January 15, 2017. So I wouldn't want to give away too many hints before then.

But just as a modest suggestion, I think that you might approach this problem along more naive lines than those that you suggest. If you want to minimise a positive integer then the smallest candidates are $1, 2, 3, \ldots $. So think about whether there is a possible solution with say $c=1$. If not , then how about $c=2,$ $c=3, \ldots$?
 
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