# How to solve such a quadratic question: If $a^2 -a -3=0$ then $a^3$ equals $a+1$, $2a+1$, $4a+1$, $4a+3$ or $5a+3$ [closed]

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.

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

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• 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. – delivosa Aug 29 '17 at 15:55
• No, you don't need the roots. if you multiply by $a$ you get $a^3=a^2+3a$. Now, can you rewrite $a^2$? – lulu Aug 29 '17 at 15:56
• Oh, these things always look easy after you see them. They are often hard to spot though. – lulu Aug 29 '17 at 16:00
• 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. – Simply Beautiful Art Aug 29 '17 at 20:56
• 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! – Simply Beautiful Art Aug 29 '17 at 20:56

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.

\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}

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.

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}

• 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? – delivosa Aug 29 '17 at 17:13
• @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. – Bill Dubuque Aug 29 '17 at 17:14
• Okay, thank you for that insight. Thought it was some shortcut you just know, still very new to this. – delivosa Aug 29 '17 at 17:15