Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $$ax + by = 7$$ $$ax^2 + by^2 = 49$$ $$ax^3 + by^3 = 133$$ $$ax^4 + by^4 = 406$$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.

share|cite|improve this question
$a,b,x,y \in \mathbb{R}$ , right, or are they integers? – cirpis Jun 5 '14 at 12:57
Should be an integer. – snivysteel Jun 5 '14 at 13:11
What have you tried so far? – user88595 Jun 5 '14 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 '14 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 '14 at 13:28

2 Answers 2

up vote 10 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 $$ 2014(x+y-xy)-100(a+b)=5956. $$

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

share|cite|improve this answer
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 '14 at 15:37
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 '14 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 '14 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 '14 at 20:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.