Year: 2007
Paper: 1
Question Number: 6
Course: LFM Stats And Pure
Section: Polynomials
There were significantly more candidates attempting this paper this year (an increase of nearly 50%), but many found it to be very difficult and only achieved low scores. In particular, the level of algebraic skill required by the questions was often lacking. The examiners' express their concern that this was the case despite a conscious effort to make the paper more accessible than last year's. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many good starts to questions soon became unstuck after a simple slip. Graph sketching was usually poor: if future candidates wanted to improve one particular skill, they would be well advised to develop this. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was pleasing to note that the applied questions were more popular this year, and many candidates scored well on at least one of these. It was however surprising how rarely answers to questions such as 5, 9, 10, 11 and 12 began with a diagram. However, the examiners were left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides. Further, and fuller, discussion of the solutions to these questions can be found in the Hints and Answers document.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1489.2
Banger Comparisons: 3
\begin{questionparts}
\item
Given that
$x^2 - y^2 = \left( x - y \right)^3$
and that $x-y = d$ (where $d \neq 0$),
express each of $x$ and $y$ in terms of $d$.
Hence find a pair of integers $m$ and $n$
satisfying $m-n = \left( \sqrt {m} - \sqrt{n} \right)^3$
where $m > n > 100$.
\item
Given that $x^3 - y^3 = \left( x - y \right)^4$
and that $x-y = d$ (where $d \neq 0$),
show that $3xy = d^3 - d^2$. Hence show that
\[
2x = d \pm d \sqrt {\frac{4d-1 }{3}}
\]
and determine a pair of distinct positive integers $m$ and $n$
such that $m^3 - n^3 = \left( m - n \right)^4$.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& x^2-y^2 &=(x-y)^3 \\
\Rightarrow && x+y &=d^2 \\
&& x-y &= d \\
\Rightarrow && x &= \tfrac12(d^2+d) \\
&& y &= \tfrac12(d^2-d)
\end{align*}
Therefore consider $x^2 = m, y^2 = n$, so $m = \tfrac14(d^2+d)^2, n = \tfrac14(d^2-d)^2$ so we want $d^2-d > 20$, so $d = 6, n = 225, m = 441$.
\item $\,$
\begin{align*}
&& x^3-y^3 &= (x-y)^4 \\
\Rightarrow && x^2+xy+y^2 &= (x-y)^3 \\
&& d^3 &= (x-y)^2+3xy \\
&& d^3 &= d^2 + 3xy \\
\Rightarrow && 3xy &= d^3 - d^2 \\
\Rightarrow && 3x(x-d) &= d^3-d^2 \\
\Rightarrow && 0 &= 3x^2-3dx-(d^3-d^2) \\
\Rightarrow && 2x &=d \pm \sqrt{d^2+4\frac{(d^3-d^2)}{3}} \\
&&&= d \pm d \sqrt{\frac{3+4d-4}{3}} \\
&&&= d \pm d \sqrt{\frac{4d-1}{3}}
\end{align*}
Therefore we need $\frac{4d-1}{3}$ to be an odd square. $y = x-d = -\frac{d}{2} \pm \frac{d}{2} \sqrt{\frac{4d-1}{3}}$. Since we want positive values, we should take the positive square roots.
$d = \frac{3 \cdot 3^2 + 1}{4} = 7$ we have $2x = 7 +7 \cdot 3 = 28 \Rightarrow x = 14, y = 7$
\end{questionparts}
This was a popular, straightforward question, which was often answered well. However, algebraic errors still occurred, for example when expanding (x – y)3.