## Saturday, May 16, 2015

### Olympiad Algebra Problem: Evaluate $ax^5+by^5$

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 $ax^5+by^5$.

Aww...this Olympiad algebra problem could be solved in at least 3 ways, therefore it is deemed to be a good Olympiad contest problem. When the opportunity of solving good problem arises, we won't say no, we will jump on the opportunity to take the time and concentrated efforts to make the most of it.

For me, I would say the most ingenious method out of the other 3 it the one to use whatever we're given wisely, with the hope that the progress won't get messy along the way.

We could make full use of the expressions in the LHS for the given 4 equations by creating new equations as follows:

$(ax+by)(\text{something else})$

That something else has to be "meaningful", and meant to help but not create more unwanted terms.

The options that we have now are

1. $a+b$

2. $x+y$

3. $ax^2+by^2$ or $ax^3+by^3$ or $ax^4+by^4$

But we can simply cross the first option out, as it will introduce the term with higher powers of $a$ and $b$ that couple with their $x$ and $y$ variables, which we don't want.

We might want to try with our second option:

$(ax+by)(x+y)=ax^2+axy+bxy+by^2$

$(ax+by)(x+y)=ax^2+by^2+xy(a+b)$---(1)

Plug in the values for $ax+by=7$ and $ax^2+by^2=49$ into (1), we get:

$7(x+y)=49+xy(a+b)$---(2)

This convinced us to continue in such a manner and to multiply $ax^2+by^2$ again with $x+y$:

$(ax^2+by^2)(x+y)=ax^3+ax^2y+bxy^2+by^3$

$(ax^2+by^2)(x+y)=ax^3+by^3+xy(ax+by)$---(3)

Plug in the values for $ax+by=7$, $ax^2+by^2=49$ and $ax^3+by^3=133$ into (3), we get:

$49(x+y)=133+xy(7)$

$7(x+y)=19+xy$---(4)

There is no way we could figure out the value for $ax^5+by^5$ yet, as the equations below are not enough to lead us to the answer.

[MATH]\color{yellow}\bbox[5px,purple]{7(x+y)=49+xy(a+b)}[/MATH]---(2) and [MATH]\color{yellow}\bbox[5px,green]{7(x+y)=19+xy}[/MATH]---(4)

Therefore, we need to multiply $ax^3+by^3$ again with $x+y$:

$(ax^3+by^3)(x+y)=ax^4+ax^3y+bxy^3+by^4$

$(ax^3+by^3)(x+y)=ax^4+by^4+xy(ax^2+by^2)$---(5)

Plug in the values for $ax^2+by^2=49$, $ax^3+by^3=133$ and $ax^4+by^4=406$ into (3), we get:

[MATH]\color{yellow}\bbox[5px,blue]{19(x+y)=58+xy(7)}[/MATH]---(6)

Aha! Now, with the equations (4) and (6), we could solve for the values for $x+y$ and $xy$:

[MATH]\color{yellow}\bbox[5px,green]{7(x+y)=19+xy}[/MATH]

[MATH]\color{yellow}\bbox[5px,green]{7(7(x+y))=7(19)+xy(7)}[/MATH]

[MATH]\color{yellow}\bbox[5px,blue]{19(x+y)=58+xy(7)}[/MATH]

$(7(7)-19)(x+y)=7(19)-58$

$x+y=2.5$

$\therefore 7(2.5)=19+xy\implies xy=-1.5$.

We are not done yet, we have to look for the value of $ax^5+by^5$, so we need to multiply $ax^4+by^4$ with $x+y$:

$(ax^4+by^4)(x+y)=ax^5+ax^4y+bxy^4+by^5$

$(ax^4+by^4)(x+y)=ax^5+by^5+xy(ax^3+by^3)$

Therefore

$(406)(2.5)=ax^5+by^5+(-1.5)(133)$

This gives

$ax^5+by^5=1214.5$