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If $ax + by = 7$, $ax^2 + by^2 = 49$, $ax^3 + by^3 = 133$, and $ax^4 +$ $by^4 = 406$, find the value of $2014(x+y-xy) - 100(a+b)$.

I came across this question in a Math Olympiad Competition and I am not sure how to solve it. Can anyone help? Thanks.

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$a,b,x,y \in \mathbb{R}$ , right, or are they integers? –  cirpis Jun 5 at 12:57
    
Should be an integer. –  snivysteel Jun 5 at 13:11
    
What have you tried so far? –  user88595 Jun 5 at 13:15
    
I tried $a^2x^2 + 2axby + b^2y^2 = ax^2 + by^2$ but I am not sure on how to carry on from this step –  snivysteel Jun 5 at 13:18
    
You have posted quite a number of contest math questions recently, do you have a link to this problem set? –  dtldarek Jun 5 at 13:28

2 Answers 2

up vote 8 down vote accepted

From $ax+by=7$, we have $ax=7-by, by=7-ax$. Noting $$ ax^2+by^2=x\cdot ax+y\cdot by=x(7-by)+y(7-ax)=7(x+y)-(a+b)xy, $$ from $ax^2+by^2=49$, we obtain $$ \tag{$*$} 7(x+y)-(a+b)xy=49. $$ Similarly, $$ ax^2=49-by^2,by^2=49-ax^2, ax^3=133-by^3,by^3=133-ax^3, $$ from which, we have $$ ax^3+by^3=x\cdot ax^2+y\cdot by^2=x(49-by^2)+y(49-ax^2)=49(x+y)-(ax+by)xy $$ and hence $49(x+y)-7xy=133$ or $$ \tag{$**$} 7(x+y)-xy=19. $$ Finally, $$ ax^4+by^4=x\cdot ax^3+y\cdot by^3=x(133-by^3)+y(133-ax^3)=133(x+y)-(ax^2+by^2)xy=133(x+y)-49xy $$ and hence $133(x+y)-49xy=406$ or $$\tag{$***$}\quad\quad\quad 19(x+y)-7xy=58. $$ From $(*), (**), (***)$, it is easy to see $$x+y=\frac{5}{2},xy=-\frac{3}{2},a+b=21 $$ and hence $$ 2014(x+y-xy)-100(a+b)=5956. $$

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Is $v=xy$? ${}{}{}{}$ –  Américo Tavares Jun 5 at 13:56
    
What is v? Why is it that it suddenly appears at the bottom. –  snivysteel Jun 5 at 14:09
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@AméricoTavares, yes. Thank you for pointing this typo and I corrected it already. –  xpaul Jun 5 at 14:09
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@snivysteel, this is a typo and I already corrected it. Thank you for telling me this. –  xpaul Jun 5 at 14:10
    
+1, Very nice answer. –  Américo Tavares Jun 5 at 15:09

These equations are in the form of the first term of the general solution of a linear recurrence with constant coefficients of order $2$. The general term of the solution is $$ u_n=ax^n+by^n $$ and the recurrence is $$ u_{n+1}=cu_n+du_{n-1} $$ Knowing two triples from the four terms $(u_1,u_2,u_3,u_4)=(7,49,133,406)=7\cdot(1,7,19,58)$ of the sequence allows to establish equations for $c$ and $d$ \begin{align} 19&=7c+d\\ 58&=19c+7d\\ \text{and consequently}&\\ 75&=30c\\ c&=\frac52\\ d&=19-7c=\frac32 \end{align} so that $$ 2u_{n+2}-5u_{n+1}-3u_n=0 $$ For the roots $x$ and $y$ of the characteristic equation we get $x+y=\frac52$ and $xy=-\frac32$. The expression $a+b=u_0$ is the term of the sequence before the given ones, $$ a+b=\frac13(2u_2-5u_1)=\frac13(98-35)=21 $$ This is sufficient to compute the crazy combination that is requested as answer.

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I like your answer, particularly because it aligns with my intuitions. Using asymptotical approach one can guess that the maximum of $|x|$ and $|y|$ is approximated by $\frac{406}{133}\simeq 3.05 \approx 3$ and its coefficient by $\frac{406}{3^4}\simeq 5.01 \approx 5$. –  dtldarek Jun 5 at 15:37
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But you can also solve $0=2x^2-5x-3=2((x-\frac54)^2-\frac32-\frac{25}{16})$ to get the solutions $x=\frac54\pm\frac74=\{3,-\frac12\}$. –  LutzL Jun 5 at 15:45
    
It is worth noting that the reason why the expression $ax^n + by^n$ satisfies such a linear recurrence relation is because of the identity $$ax^{n+1} + by^{n+1} = (x+y)(ax^n + by^n) - (xy)(ax^{n-1} + by^{n-1}).$$ –  heropup Jun 5 at 19:44
    
Yes, this is one formulation of it. Using generating functions and partial fraction decomposition gives another derivation of this identity. –  LutzL Jun 5 at 20:34

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