Year: 2010
Paper: 1
Question Number: 8
Course: LFM Pure
Section: Proof
There were significantly more candidates attempting this paper than last year (just over 1000), and the scores were much higher than last year (presumably due to the easier first question): fewer than 2% of candidates scored less than 20 marks overall, and the median mark was 61. The pure questions were the most popular as usual, though there was much more variation than in some previous years: questions 1, 3, 4 and 6 were the most popular, while question 7 (on vectors) was intensely unpopular. About half of all candidates attempted at least one mechanics question, and 15% attempted at least one probability question. The marks were unsurprising: the pure questions generally gained the better marks, while the mechanics and probability questions generally had poorer marks. A sizeable number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have often restricted my attention below to those attempts which counted as one of the six highest-scoring answers, and referred to these as "significant attempts". The majority of candidates did begin with question 1 (presumably as it appeared to be the easiest), but some spent far longer on it than was wise. Some attempts ran to over eight pages in length, especially when they had made an algebraic slip early on, and used time which could have been far better spent tackling another question. It is important to balance the desire to finish the question with an appreciation of when to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, "Hence, or otherwise, show . . ." gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, but in Question 5 this year, the "otherwise" approach was very popular.) On some questions, some candidates tried to work forwards from the given question and backwards from the answer, hoping that they would meet somewhere in the middle. While this worked on occasion, it often required fudging. It is wise to remember that STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
\begin{questionparts}
\item Suppose that $a$, $b$ and $c$ are integers
that satisfy the equation
\[
a^{3}+3b^{3}=9c^{3}.
\]
Explain why $a$ must be divisible by 3,
and show further that both $b$ and $c$ must also be divisible by 3.
Hence show that the only integer solution is $a=b=c=0\,$.
\item Suppose that $p$, $q$ and $r$ are integers
that satisfy the equation
\[
p^4 +2q^4 = 5r^4
\,.\]
By considering the possible final digit of each
term, or otherwise, show that
$p$ and $q$ are divisible by 5. Hence
show that the only integer solution is $p=q=r=0\,$.
\end{questionparts}\begin{questionparts}
\item Since $a^3 = 9c^3 - 3b^3 = 3(3c^3-b^3)$ we must have $3 \mid a^3$. But since $3$ is prime, $3 \mid a$. Since $3 \mid a$ we can write $a = 3a'$ for some $a' \in \mathbb{Z}$. Therefore our equation is $27(a')^3 + 3b^3 = 9c^3 \Rightarrow 9(a')^3 + b^3 = 3c^3$ which means that $3 \mid b$ by the same argument from earlier. So $b = 3b'$ so the equation is $9(a')^3 + 27(b')^3 = 3c^3 \Rightarrow 3(a')^3 + 9(b')^3 = c^3$ which means that $3 \mid c$.
Suppose $(a,b,c)$ is the smallest measured by $a^2+b^2+c^2$ with $a, b, c\neq 0$. Then $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$ is also a solution. But this contradicts that we had found the smallest solution. Therefore the only possible solution is $(0,0,0)$ which clearly works.
\item Consider $p, q \pmod{5}$. By $FLT$ $p^4, q^4 = 0, 1 \pmod{5}$ so $p^4+2q^4 \in \{0, 1, 2, 3\}$ and in particular the only way they are divisible by $5$ is if $p \equiv q \equiv 0 \pmod{5}$. Therefore $p = 5p', q = 5q'$ and so $5^4(p')^4 + 5^4(q')^4 = 5r^4 \Rightarrow r^4 = 5(25(p')^4 + 25(q')^4) \Rightarrow 5\mid r^4 \Rightarrow 5 \mid r$. Therefore we can use the same argument about the smallest solution to show that $p = q= r = 0$
\end{questionparts}
This was a moderately popular question and candidates obtained a broad spread of marks on it. For the first part, almost all candidates successfully argued that a³ is divisible by 3, but a significant number could not give a reasonably convincing explanation for why a itself is divisible by 3. We did not require a perfect argument, but there had to be a mention of 3 being prime. It was common to see this either asserted ("since 3 | a³, then 3 | a") or, less commonly, something creative like: "a = ∛(a³) = ∛(3(3c³ − b³)), and as a is an integer, a must be divisible by 3". A good number of candidates went on to correctly explain why b and c were divisible by 3, but quite a few talked about b³ = 3c³ − ⅓a³ and c³ = ⅑a³ + ⅓b³ being divisible by 3 without any justification for these assertions; the idea of writing a multiple of 3 as 3k for some k was appreciated by some candidates but overlooked by others. A few candidates also made basic algebraic errors when rearranging the equation a³ + 3b³ = 9c³; more care is needed! It was nice to see that a fair number of candidates knew about modular arithmetic and could use it to construct effective arguments both here and in part (ii). A number of candidates showed some serious misconceptions about divisibility, with a common error being 3 | (r + s) ⇒ 3 | r and 3 | s. A few expanded ∛(a³ + b³) as ∛a + ∛b, and there were even occurrences of r + s = t ⇒ r = t/s. Other somewhat common failings were attempting to prove general statements by using particular examples (such as "take a = 42"), making unsubstantiated assertions (for example "since a³/3 is an integer, so is a³/9"), and misuse of "similarly": just because 27 | a³, it does not follow "similarly" that 27 | b³ and 27 | c³ unless it has already been shown that 3 | b and 3 | c. The final step of the argument, the infinite descent argument, was poorly understood and served to differentiate the strongest candidates from the rest. Again, we did not expect a perfect argument, but merely some explanation that a non-zero solution would give rise to another, smaller, non-zero solution, and so on, which is impossible. For part (ii), most candidates who attempted it realised that there were very few possibilities for the final digits of fourth powers, though a few went to the effort of calculating 8⁴ and 9⁴ explicitly rather than just considering final digits. Some thought that 5r⁴ could only end in 5, forgetting the possibility that it might end in 0. Nevertheless, many understood what they were meant to do and tried to argue that p and q must both be multiples of 5. Some succeeded, but others made errors in their logic and did not consider all possible cases, for example, many candidates considered the case 5 ∤ p but not 5 | p. Others effectively said "if 5 | p, then we must have 5 | q," but did not show that we must have 5 | p. Finally, as in part (i), there were some who argued that 5 | r⁴ but did not explain how to deduce that 5 | r, and many who were stuck on the final proof by infinite descent. It was perhaps unsurprising that so many candidates appealed to Fermat's Last Theorem in their attempts to prove these results, even though it was not at all relevant to the question. (It was equally unsurprising that there was almost no use of Fermat's Little Theorem to help with the calculations required in part (ii) by stating that a⁴ ≡ 1 (mod 5) whenever 5 ∤ a.)