One Worked Example of Solving Heuristic IMO Math Problem

Olympiad Algebra Problem:

If $a,\,b,\,x,\,y\in R$ such that

$ax +by =7$

$ax^2+by^2=49$

$ax^3+by^3=133$

$ax^4+by^4=406$

Evaluate $ax^5+by^5$.

The solution from the U.K. mathematician:

He thought of solving this problem by making full use of the given four equations by mix-matching them so that he could get something that looks like the following:

$(\text{something})(ax^5+by^5)=(\text{something else_1})(\text{something else_2})\pm(\text{something else_3})(\text{something else})$---(*)

with the condition all those something else must come from $ax +by$, $ax^2+by^2$, $ax^3+by^3$ and/or $ax^4+by^4$.

There is something that could guide us so we don't sway too far from the correct identity, that is the powers for each term, if we let the LHS of the equation (*) be

$(\text{something})(ax^5+by^5)=(ax^2+by^2)^2(ax^5+by^5)$

Then we know the sum of the exponents in each term from both sides of the equation (*) adds up to 9.

After doing some experiments by trial and error, it's not hard but merely required some extra work to find that:

[MATH]\color{yellow}\bbox[5px,purple]{((ax^2+by^2)^2-(ax+by)(ax^3+by^3))(ax^5+by^5)}[/MATH]

[MATH]\color{black}{=}\color{yellow}\bbox[5px,green]{2(ax^2+by^2)(ax^3+by^3)(ax^4+by^4)}\color{black}{\,-}\color{yellow}\bbox[5px,blue]{(ax^3+by^3)^3}\color{black}{\,-}\color{black}\bbox[5px,orange]{(ax+by)(ax^4+by^4)^2}[/MATH]

since both LHS and RHS give us the exact same expression: $-abxy(ax^5+by^5)(x^2-2xy+y^2)$.

Therefore,

[MATH]\color{yellow}\bbox[5px,purple]{((49)^2-(7)(133))(ax^5+by^5)}\color{black}{=\,\,}\color{yellow}\bbox[5px,green]{2(49)(133)(406)}\color{black}{\,\,-\,\,}\color{yellow}\bbox[5px,blue]{(133)^3}\color{black}{\,\,-\,\,}\color{black}\bbox[5px,orange]{(7)(406)^2}[/MATH]

$1470(ax^5+by^5)=1785315$

$ax^5+by^5=\dfrac{1785315}{1470}=1214.5$

and we are hence done.


Okay, I will admit it, it's doubly hard in figuring out what combinations of the terms from the mix-matching that finally works. Yes, but, I am fairly certain you will agree with me this problem is related to not taking no for an answer, perseverance, being creative and thinking outside the box. And it is so important for the math educators to engage students in the practice of critical thinking. Great leaders/bosses are never satisfied with

• traditional practice, 
• static thinking, 
• conventional wisdom, or 
• common performance. 

Therefore, it's time to help student to shift their thinking style. After all, it takes practice, hard work, and positive thinking to succeed.

When you finally figure out 

[MATH]\color{yellow}\bbox[5px,purple]{((ax^2+by^2)^2-(ax+by)(ax^3+by^3))(ax^5+by^5)}[/MATH]

[MATH]\color{black}{=}\color{yellow}\bbox[5px,green]{2(ax^2+by^2)(ax^3+by^3)(ax^4+by^4)}\color{black}{\,-}\color{yellow}\bbox[5px,blue]{(ax^3+by^3)^3}\color{black}{\,-}\color{black}\bbox[5px,orange]{(ax+by)(ax^4+by^4)^2}[/MATH]

and find that it actually works out so perfectly well, don't you feel thoroughly satisfied?