Year: 2011
Paper: 2
Question Number: 2
Course: LFM Pure
Section: Proof
There were just under 1000 entries for paper II this year, almost exactly the same number as last year. After the relatively easy time candidates experienced on last year's paper, this year's questions had been toughened up significantly, with particular attention made to ensure that candidates had to be prepared to invest more thought at the start of each question – last year saw far too many attempts from the weaker brethren at little more than the first part of up to ten questions, when the idea is that they should devote 25-40 minutes on four to six complete questions in order to present work of a substantial nature. It was also the intention to toughen up the final "quarter" of questions, so that a complete, or nearly-complete, conclusion to any question represented a significant (and, hopefully, satisfying) mathematical achievement. Although such matters are always best assessed with the benefit of hindsight, our efforts in these areas seem to have proved entirely successful, with the vast majority of candidates concentrating their efforts on four to six questions, as planned. Moreover, marks really did have to be earned: only around 20 candidates managed to gain or exceed a score of 100, and only a third of the entry managed to hit the half-way mark of 60. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were attempted by almost all candidates; 3 and 4 by around three-quarters of them; 6, 7 and 9 by around half; the remaining questions were less popular, and some received almost no "hits". Overall, the highest scoring questions (averaging over half-marks) were 1, 2 and 9, along with 13 (very few attempts, but those who braved it scored very well). This at least is indicative that candidates are being careful in exercising some degree of thought when choosing (at least the first four) 'good' questions for themselves, although finding six successful questions then turned out to be a key discriminating factor of candidates' abilities from the examining team's perspective. Each of questions 4-8, 11 & 12 were rather poorly scored on, with average scores of only 5.5 to 6.6.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Write down the cubes of the integers $1, 2, \ldots , 10$.
The positive integers $x$, $y$ and $z$, where $x < y$, satisfy
\[
x^3+y^3 = kz^3\,,
\tag{$*$}
\]
where $k$ is a given positive integer.
\begin{questionparts}
\item
In the case $x+y =k$, show that
\[
z^3 = k^2 -3kx+3x^2\,.
\]
Deduce that $(4z^3 - k^2)/3$ is a perfect square
and that $\frac14 {k^2} \le z^3 < k^2\,$.
Use these results to find a solution of $(*)$ when $k=20$.
\item By considering the case $x+y = z^2$,
find two solutions of $(*)$ when
$k=19$.
\end{questionparts}\begin{array}{c|c}
n & n^3 \\ \hline
1 & 1 \\
2 & 8 \\
3 & 27 \\
4 & 64 \\
5 & 125 \\
6 & 216 \\
7 & 343 \\
8 & 512 \\
9 & 729 \\
10 & 1000 \\
\end{array}
\begin{questionparts}
\item $\,$
\begin{align*}
&& x^3 + y^3 &= kz^3 \\
\Rightarrow &&k(x^2-xy+y^2)&=kz^3 \\
\Rightarrow && z^3 &= (x+y)^2-3xy \\
&&&= k^2-3x(k-x) \\
&&&= k^2-3xk+3x^2 \\
\\
\Rightarrow && \frac{4z^3-k^2}{3} &= \frac{4(k^2-3xk+3x^2)-k^2}{3} \\
&&&= \frac{3k^2-12xk+12x^2}{3} \\
&&&= k^2-4xk+4x^2 \\
&&&= (k-2x)^2
\end{align*}
Therefore $\frac{4z^3-k^2}{3}$ is a perfect square and so $4z^3 \geq k^2 \Rightarrow z^3 \geq \frac14k^2$. Clearly $kz^3 < x^3+3x^2y+3xy^2+y^3 = k^3 \Rightarrow z^3 < k^2$, therefore $\frac14 k^2 \leq z^3 < k^2$
Therefore if $k = 20$, $100 \leq z^3 < 400 \Rightarrow z \in \{ 5, 6,7\}$. Mod $3$ it is clear that $4z^3-k^2$ is not divisible by $3$ for $z = 5,6$ therefore $z = 7$
\begin{align*}
&& 343 &= 3x^2-60x+400 \\
\Rightarrow && 0 &= 3x^2-60x+57 \\
\Rightarrow && 0 &= x^2-20x+19 \\
\Rightarrow && x &= 1,19
\end{align*}
Therefore a solution is $1^3 + 19^3 = 20 \cdot 7^3$
\item When $x+y = z^2$ we must have
\begin{align*}
&& x^3 + y^3 &= kz^3 \\
\Rightarrow &&(x^2-xy+y^2)&=kz \\
\Rightarrow && kz &= (x+y)^2-3xy \\
&&&= z^4-3x(z^2-x)\\
&&&= z^4-3xz^2+3x^2 \\
\Rightarrow && 0 &= 3x^2-3z^2x+z^4-kz \\
\\
\Rightarrow && 0 &\leq \Delta = 9z^4-12(z^4-kz) \\
&&&=12kz-3z^4 \\
\Rightarrow && z^3 &\leq 4k
\end{align*}
If $k = 19$ this means $z \leq 4$
\begin{array}{c|c|c|c}
z & 19z^3 & x & y \\ \hline
1 & 19 & - & - \\
2 & 152 & 3 & 5 \\
3 & 513 & 1 & 8
\end{array}
So two solutions are $1^3+8^3 = 19 \cdot 3^3$ and $3^3+5^3=19 \cdot 2^3$
\end{questionparts}
Personally, this was my favourite question, even though it was ultimately (marginally) deflected from its original purpose of expressing integers as sums of two rational cubes. Given that the question explicitly involves inequalities (which are, as a rule, never popular) and cubics rather than quadratics, it was slightly surprising to find that it was the most popular question on the paper. However, although the average score on the question was almost exactly 10, these two issues then turned out to be the biggest stumbling-blocks to a completely successful attempt as candidates progressed through the question, both in establishing the given inequalities and then in the use of them. In particular, it was noted that many candidates "proved" the given results by showing that they implied something else that was true, rather than by deducing them from something else known to be true; such logical flaws received little credit in terms of marks. The purpose of this preliminary work was to enable the candidates to whittle down the possibilities to a small, finite list and then provide them with some means of testing each possibility's validity. This help was often ignored in favour of starting again. In general, though, part (i) was done reasonably well; as was (ii) by those who used (i)'s methodology as a template. Only a very few candidates were bold enough to attempt (ii) successfully without any reference to (i)'s methods; indeed, this arithmetic approach was how the question was originally posed (as part (i), of course) before proceeding onto the algebra. Noting that the wording of the question does not demand any particular approach in order to find the required two solutions to the equation x³ + y³ = 19z³, a reasonably confident arithmetician might easily note that 19 × 2³ = 152 = 3³ + 5³ and 19 × 3³ = 513 = 1³ + 8³ and it isn't even necessary to look very far for two solutions. For 10 marks, this is what our transatlantic cousins would call "a steal".