Year: 2022
Paper: 3
Question Number: 2
Course: UFM Additional Further Pure
Section: Number Theory
No solution available for this problem.
One question was attempted by well over 90% of the candidates two others by about 90%, and a fourth by over 80%. Two questions were attempted by about half the candidates and a further three questions by about a third of the candidates. Even the other three received attempts from a sixth of the candidates or more, meaning that even the least popular questions were markedly more popular than their counterparts in previous years. Nearly 90% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Suppose that there are three non-zero integers $a$, $b$ and $c$ for which $a^3 + 2b^3 + 4c^3 = 0$. Explain why there must exist an integer $p$, with $|p| < |a|$, such that $4p^3 + b^3 + 2c^3 = 0$, and show further that there must exist integers $p$, $q$ and $r$, with $|p| < |a|$, $|q| < |b|$ and $|r| < |c|$, such that $p^3 + 2q^3 + 4r^3 = 0$. Deduce that no such integers $a$, $b$ and $c$ can exist.
\item Prove that there are no non-zero integers $a$, $b$ and $c$ for which $9a^3 + 10b^3 + 6c^3 = 0$.
\item By considering the expression $(3n \pm 1)^2$, prove that, unless an integer is a multiple of three, its square is one more than a multiple of $3$. Deduce that the sum of the squares of two integers can only be a multiple of three if each of the integers is a multiple of three.
Hence prove that there are no non-zero integers $a$, $b$ and $c$ for which $a^2 + b^2 = 3c^2$.
\item Prove that there are no non-zero integers $a$, $b$ and $c$ for which $a^2 + b^2 + c^2 = 4abc$.
\end{questionparts}
The fifth most popular question, being attempted by just a little over half the candidates, it was the fourth most successful with a mean score of 9/20. Whilst the algebra associated with this question was not difficult, the logic and communication required was certainly too much for many students. In part (i) it required some justification that a had to be even. Contradiction or infinite descent could be used but either way the argument had to be made clear. Claiming "this can be continued forever" or moduli were always decreasing would eventually get to zero was not good enough. Successful candidates were able to explain why the integer nature of the solutions was vital to reach a contradiction. In part (ii) many candidates were able to see that this was a similar problem to the first one, and most observed that divisibility by three was now the key idea. In part (iii), many candidates were able to consider the remainders when divided by 3, but again many struggled to communicate clearly an argument leading to the final contradiction. By part (iv) most candidates were expecting to recreate the original equation again and the fact that this did not happen meant some came to a dead halt. Other were either oblivious to the issue or were bluffing their way through as a slightly more subtle argument was now required.