Review question

# If $x^3 = 2x+1$, what is $x^k$ as a quadratic? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8586

## Solution

Suppose that $x$ satisfies the equation $\begin{equation*} x^3=2x+1.\label{eq:star}\tag{*} \end{equation*}$
1. Show that $x^4=x+2x^2\qquad\text{and}\qquad x^5=2+4x+x^2.$

If we multiply $\eqref{eq:star}$ by $x$, we obtain the first of the two equations.

For the second one, we multiply $\eqref{eq:star}$ by $x^2$ to get $x^5=2x^3+x^2$. We can now substitute $x^3$ using $\eqref{eq:star}$, and the equation becomes $x^5=2(2x+1)+x^2=x^2+4x+2.$

1. For every integer $k\geq0$, we can uniquely write $x^k=A_k+B_kx+C_kx^2$ where $A_k$, $B_k$, $C_k$ are integers. So, in part (i), it was shown that $A_4=0,\ B_4=1,\ C_4=2 \qquad\text{and}\qquad A_5=2,\ B_5=4,\ C_5=1.$ Show that $A_{k+1}=C_k,\qquad B_{k+1}=A_k+2C_k,\qquad C_{k+1}=B_k.$

Since $x^k=A_k+B_kx+C_kx^2$, we can multiply this by $x$ to obtain $x^{k+1}=A_kx+B_kx^2+C_kx^3.$ We can then use $\eqref{eq:star}$ to rewrite $x^3$ as $2x+1$, and thereby obtain (after collecting like terms): $x^{k+1}=C_k+(A_k+2C_k)x+B_kx^2.$ But we also have, from the statement of the question, that $x^{k+1}=A_{k+1}+ B_{k+1}x+C_{k+1}x^2.$ Since the coefficients in the two ways of writing out $x^{k+1}$ must be equal, we get $A_{k+1}=C_k$, $B_{k+1}=A_k+2C_k$ and $C_{k+1}=B_k$ as required.

1. Let $D_k=A_k+C_k-B_k.$ Show that $D_{k+1}=-D_k$ and hence that $A_k+C_k=B_k+(-1)^k.$

Using the results from the previous part and the fact that $D_{k+1}=A_{k+1}+C_{k+1}-B_{k+1}$, we get $D_{k+1}=C_k+B_k-(A_k+2C_k)=-(A_k+C_k-B_k)=-D_k.$

Rearranging the equation $D_k=A_k+C_k-B_k$, we get $A_k+C_k=B_k+D_k$. Therefore we now need to show that $D_k=(-1)^k$.

We will prove by induction that $D_k=(-1)^k$, for every $k\geq3$.

For $k=3$, $A_3=1$, $B_3=2$, $C_3=0$, so $D_3=1+0-2=-1=(-1)^3$, so the result holds.

If the result holds for $k$, then, as $D_{k+1}=-D_k$, $D_{k+1}=-(-1)^k=(-1)^{k+1}.$

So, by mathematical induction, $D_k=(-1)^k$, and therefore $A_k+C_k=B_k+(-1)^k$, as required.

1. Let $F_k=A_{k+1}+C_{k+1}$. Show that $F_k+F_{k+1}=F_{k+2}.$

We will rewrite $F_{k+1}$ and $F_{k+2}$ in terms of $A_{k+1}$, $B_{k+1}$ and $C_{k+1}$, using the results of part (ii) repeatedly.

\begin{align} &F_{k+1}=A_{k+2}+C_{k+2}=C_{k+1}+B_{k+1} \label{eq:1}\\ &F_{k+2}=A_{k+3}+C_{k+3}=C_{k+2}+B_{k+2}= B_{k+1}+A_{k+1}+2C_{k+1}\label{eq:2} \end{align}

By $\eqref{eq:1}$, $F_k+F_{k+1}=A_{k+1}+C_{k+1}+C_{k+1}+B_{k+1}= A_{k+1}+B_{k+1}+2C_{k+1},$ so, by equation $\eqref{eq:2}$, $F_k+F_{k+1}=F_{k+2}.$