Integral Roots of a Cubic Equation | AIME 2011 Problem 15

#MathOlympiad #Polynomials #Algebra

Here is the solution to AIME 2011 Problem 15!

4:22 Remark: Before picking the right version of "a^2+ab+b^2 = 2011" to consider, we need to look at the size of the three numbers. For a,b,c, we have at least 1 positive number, and let a be that number. Then b and c are roots to the equation x^2 + ax + a^2 - 2011, and both roots are negative, so b,c are both negative.

This makes b^2+bc+c^2 to be less than (b+c)^2.

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