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enter image description here

This is a Belgian math olympiad question. https://www.vwo.be/vwo/files/2r2017.pdf

Sorry, the picture is in Dutch I will translate it:

If $a^2 -a -3=0 $ then $a^3$ is equal to:

(A) $a+1$ (B) $2a+1$ (C) $4a+1$ (D) $4a+3$ (E) $5a+3$

I've tried $a^2 = a + 3$ then multiply by $a$. Doesn't work.

I've tried finding the roots in the standard manners ($x_1 + x_2 = -b/a$ and $x_1x_2 = c/a$ and also the quadratic formula) but of course that doesn't work as they ask it in terms of $a$ and I can't figure it out from the roots directly, so I will be trying your complete algebra approach thanks.

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closed as off-topic by Théophile, Henrik, Simply Beautiful Art, José Carlos Santos, Did Aug 29 '17 at 21:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Théophile, Henrik, Simply Beautiful Art, José Carlos Santos, Did
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ I've tried $a^2 = a + 3$ then multiply by a. Doesn't work. I've tried finding the roots in the standard manners (x1 + x2 = -b/a and x1*x2 = c/a and also the quadratic formula) but of course that doesn't work as they ask it in terms of a and i can't figure it out from the roots directly, so I will be trying your complete algebra approach thanks. $\endgroup$ – delivosa Aug 29 '17 at 15:55
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    $\begingroup$ No, you don't need the roots. if you multiply by $a$ you get $a^3=a^2+3a$. Now, can you rewrite $a^2$? $\endgroup$ – lulu Aug 29 '17 at 15:56
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    $\begingroup$ Oh, these things always look easy after you see them. They are often hard to spot though. $\endgroup$ – lulu Aug 29 '17 at 16:00
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    $\begingroup$ I've downvoted and voted to close this question because it looks very much like a homework question and is a PSQ. In the future, please include more context such as what's holding you back. $\endgroup$ – Simply Beautiful Art Aug 29 '17 at 20:56
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    $\begingroup$ Please follow the guidelines outlined by How to ask a good question? and How to ask a homework question?. Low quality questions run the risk of being closed and deleted, and repeated closures and deletions may trigger a question ban. Thank you! $\endgroup$ – Simply Beautiful Art Aug 29 '17 at 20:56
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Since the general method may not be clear from other answers it may help to describe it explicitly. Given that $a$ is a root of a polynomial $g(x)\ne 0,\,$ i.e. $\,\color{#c00}{g(a)=0},\,$ suppose we wish to compute $f(a)\,$ for some polynomial $\,f(x).\,$ By the Polynomial Division Algorithm we can divide $\,f(x)\,$ by $\,g(x)$

$$\begin{align} f(x) \,&=\, r(x)\, +\, q(x)\, g(x),\ \ {\rm with}\ \ \deg r < \deg g\\[.2em] \Rightarrow\,\ f(a) &=\, r(a)\, +\, q(a)\, \color{#c00}{g(a)}\\[.2em] &=\, r(a)\ \ \ \ \ {\rm by}\ \ \ \ \color{#c00}{g(a) = 0} \end{align}$$

Since we need only the remainder $\,r(a)\,$ it is more efficient to use arithmetic modulo $\,g(x),\,$ vs. the full-blown division algorithm, e.g. $\,g(x) = x^{\large 2}-x-3\,$ in OP $ $ so

$$\begin{align} \bmod{\,x^{\large 2}\!-\!x\!-\!3}\!:\,\ \color{#c00}{x^{\large 2}}&\equiv \,\color{#0a0}{x\,+\,3}\\[.2em] \Rightarrow\ \ x^{\large 3}&\equiv \color{#c00}{x^{\large 2}}+3x\,\ \ {\rm by}\,\ \ x * {\rm prior}\\[.2em] &\equiv \color{#0a0}{x+3}+3x\\[.2em] &\equiv 4x+3 \end{align}$$

Continuing we can rewrite $\,x^{\large 4},x^{\large 5}\ldots$ all as linear polynomials in $x$, and use this table to quickly rewrite any polynomial $f(x)$ as a linear polynomial $\, f\bmod g\,=\, r\,$ in the above notation.

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$$\begin{align}0&=a^2-a-3\\a^2&=a+3\qquad\qquad~~~(1)\\a^3&=a^2+3a\\a^3&=(a+3)+3a\qquad\text{by $(1)$}\\a^3&=4a+3\end{align}$$

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From the given equation,

$$a^2=a+3.$$

Then multiplying by $a$ and substituting,

$$a^3=a^2+3a=a+3+3a.$$

Zo simple.


The "hard" way:

The roots of the quadratic are $$a=\frac{1\pm\sqrt{13}}2.$$

Then taking the cube (by the binomial theorem), $$a^3=\frac{1\pm3\sqrt{13}+3\cdot13\pm13\sqrt{13}}8=5\pm2\sqrt{13}.$$

To account for the square root of $13$, you take $4a$, giving $2\pm2\sqrt{13}$ and add $3$ to adjust the integer term.

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Another way to approach this is to use $(a^2-a+1)(a+1)=a^3+1$. Then you can say

$$\begin{align}a^2-a-3&=0\\ a^2-a+1&=4\\ (a^2-a+1)(a+1)&=4(a+1)\\ a^3+1&=4a+4\\ a^3&=4a+3\end{align}$$

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  • $\begingroup$ Thanks, I would like to know how you can learn such "tricks". Are the Art of Problem Solving books good for these sort of things? $\endgroup$ – delivosa Aug 29 '17 at 17:13
  • $\begingroup$ @delivosa It's not a "trick". Rather, it is a special case of computing modulo a polynomial - see my answer. This is algebraically reified when one studies quotient rings. $\endgroup$ – Bill Dubuque Aug 29 '17 at 17:14
  • $\begingroup$ Okay, thank you for that insight. Thought it was some shortcut you just know, still very new to this. $\endgroup$ – delivosa Aug 29 '17 at 17:15

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