Solving 2002 Sweden Math Olympiad Question | Solving a Symmetrical System of Equations

🔴Solving 2002 Sweden Math Olympiad Question | Solving an if then question using a Symmetrical System of Equations

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In this video, we are dealing with one of the 2002 Sweden math olympiad questions, where we are given a nice symmetrical system of equations, and we are supposed to evaluate an expression.

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topics covered in this video:
system of equations
system of equations from sweden
math olympiad question
how to evaluate
if then question

#MathOlympiadQuestion #SystemOfEquations #AnonMath

[NadiehFan]

Each of the cubic equations in a and b has three solutions, two of which are complex, so the statement of the problem should have included the requirement that a and b are assumed to be real, otherwise a + b will not have a single value.

The substitutions α = a − 1 or a = α + 1 and β = b − 1 or b = β + 1 reduce the cubic equations to so-called depressed cubic equations in α and β, that is, cubic equations lacking a quadratic term. This is a standard type of linear substitution when dealing with cubic equations and it is easy to see why these particular substitutions give equations in α and β lacking a quadratic term. The sum of the roots of each of the equations in a and b is 3, so the sum of the three roots of each of the equations in α = a − 1 and β = b − 1 must be zero, implying that the coefficient of the quadratic term of these equations is zero.

To prove that

α² − αβ + β² + 2

cannot be zero for any real values of α and β you could also view this expression as a quadratic in either α or β. For any given real value of β the discriminant of this expression viewed as a quadratic in α is

(−β)² − 4·1·(β² + 2) = −3β² − 8

which is negative, so for any real value of β there exists no real value of α which will make α² − αβ + β² + 2 equal to zero. This implies that there exists no pair (α, β) of reals which will make α² − αβ + β² + 2 equal to zero.

Alternatively, we can also consider that the identity

(α − β)(α² − αβ + β²) = α³ − β³

implies that α² − αβ + β² must be positive for real α and β whenever α ≠ β since α³ − β³ > 0 for α − β > 0 and α³ − β³ < 0 for α − β < 0, that is, α − β and α³ − β³ always have the same sign if α ≠ β. Also, if α = β, then α² − αβ + β² = α² = β². So, α² − αβ + β² is never negative for real α and β, and α² − αβ + β² = 0 if and only if α = β = 0. So, we have α² − αβ + β² + 2 ≥ 2 for any real α and β.