Year: 2013
Paper: 2
Question Number: 7
Course: LFM Pure
Section: Proof
All questions were attempted by a significant number of candidates, with questions 1 to 3 and 7 the most popular. The Pure questions were more popular than both the Mechanics and the Probability and Statistics questions, with only question 8 receiving a particularly low number of attempts within the Pure questions and only question 11 receiving a particularly high number of attempts.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item Write down a solution of the equation
\[
x^2-2y^2 =1\,,
\tag{$*$}
\]
for which $x$ and $y$ are non-negative integers.
Show that, if $x=p$, $y=q$ is a solution of ($*$), then so also is $x=3p+4q$, $y=2p+3q$. Hence find two solutions of $(*)$ for which $x$ is a positive odd integer and $y$ is a positive even integer.
\item Show that, if $x$ is an odd integer and $y$ is an even integer, $(*)$ can be written in the form
\[
n^2 = \tfrac12 m(m+1)\,,
\]
where $m$ and $n$ are integers.
\item The positive integers $a$, $b$ and $c$ satisfy
\[
b^3=c^4-a^2\,,
\]
where $b$ is a prime number. Express $a$ and $c^2$ in terms of $b$ in the two cases that arise.
Find a solution of $a^2+b^3=c^4$, where $a$, $b$ and $c$ are positive integers but $b$ is not prime.
\end{questionparts}\begin{questionparts}
\item $(x,y) = (1,0)$ we have
Suppose $p^2-2q^2 = 1$, then
\begin{align*}
&& (3p+4q)^2-2\cdot(2p+3q)^2 &= 9p^2+24pq + 16q^2 - 2\cdot(4p^2+12pq+9q^2) \\
&&&= p^2(9-8) + pq(24-24) + q^2(16-18) \\
&&&= p^2 - 2q^2 = 1
\end{align*}
So we have:
\begin{array}{c|c}
x & y \\ \hline
1 & 0 \\
3 & 2 \\
17 & 12 \\
\end{array}
\item Suppose $x = 2m+1$ and $y = 2n$ then
\begin{align*}
&& 1 & = x^2 - 2y^2 \\
&&&= (2m+1)^2 - 2(2n)^2 \\
&&&= 4m^2 + 4m + 1 - 8n^2 \\
\Leftrightarrow && n^2 &= \frac{m(m+1)}{2}
\end{align*}
\item Suppose $b^3 = c^4 - a^2 =(c^2-a)(c^2+a)$, since $b$ is prime and $c^2 + a > c^2-a$ we must have:
\begin{align*}
&& p = c^2-a && p^2 =c^2 +a \\
\Rightarrow && c^2 = \frac{p+p^2}{2} && a = \frac{p^2-p}2\\
&& 1 = c^2-a && p^3 = c^2+a \\
\Rightarrow && c^2 = \frac{p^3+1}{2} && a = \frac{p^3-1}{2}
\end{align*}
Note that $c^2 = \frac{p(p+1)}{2}$ is reminicent of our first equation, so suppose $n = c = 6$ and $p = m = 8$ then
$6^4 = 8^3 + 28^2$
\end{questionparts}
This question was attempted by a large number of candidates, only slightly fewer than question 2, and was one of the more successful ones with an average score above half of the marks. While some candidates proved the converse of the required result, part (i) of the question was generally done well, although a surprising number of candidates did not write down the numerical solutions when asked. Those students who realised the way to write x and y in terms of m and n reached the result of part (ii) easily, while others sometimes spent a lot of effort on this making little or no progress. In part (iii) many candidates spotted the difference of two squares, but some did not realise that there would be two ways to factorise. Only very few students were able to solve the final part of the question.