Year: 2013
Paper: 3
Question Number: 5
Course: LFM Pure
Section: Proof
With the number of candidates submitting scripts up by some 8% from last year, and whilst inevitably some questions were more popular than others, namely the first two, 7 then 4 and 5 to a lesser extent, all questions on the paper were attempted by a significant number of candidates. About a sixth of candidates gave in answers to more than six questions, but the extra questions were invariably scoring negligible marks. Two fifths of the candidates gave in answers to six questions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1487.0
Banger Comparisons: 1
In this question, you may assume that, if $a$, $b$ and $c$ are positive integers such that $a$ and $b$ are coprime and $a$ divides $bc$, then $a$ divides $c$. (Two positive integers are said to be \textit{coprime} if their highest common factor is 1.)
\begin{questionparts}
\item Suppose that there are positive integers $p$, $q$, $n$ and $N$ such that $p$ and $q$ are coprime and $q^nN=p^n$. Show that $N=kp^n$ for some positive integer $k$ and deduce the value of $q$. Hence prove that, for any positive integers $n$ and $N$, $\sqrt[n]N$ is either a positive integer or irrational.
\item Suppose that there are positive integers $a$, $b$, $c$ and $d$ such that $a$ and $b$ are coprime and $c$ and $d$ are coprime,
and $a^ad^b = b^a c^b \,$.
Prove that $d^b = b^a$ and deduce that, if $p$ is a prime factor of $d$, then $p$ is also a prime factor of $b$. If $p^m$ and $p^n$ are the highest powers of the prime number $p$ that divide $d$ and $b$, respectively, express $b$ in terms of $a$, $m$ and $n$ and hence show that $p^n\le n$. Deduce the value of $b$. (You may assume that if $x > 0$ and $y\ge2$ then
$y^x > x$.) Hence prove that, if $r$ is a positive rational number such that $r^r$ is rational, then $r$ is a positive integer.
\end{questionparts}\begin{questionparts}
\item Suppose $q^nN = p^n$ then since $(p,q) =1$ we must have $p \mid N$, and then by dividing both $p^n$ and $N$ by $p$ we can repeat this process $n$ times to find that $N = kp^n$ and in particular $q = 1$.
Suppose $\sqrt[n]{N} = \frac{p}q$ for $p,q$ coprime positive integers (ie it is not irrational), then $q^nN = p^n$ and so $q = 1$ and in fact $\sqrt[n]{N}$ is an integer so $N$.
\item Suppose $(a,b) = 1, (c,d) = 1$ and $a^ad^b = b^ac^b$, then since $(a,b) = 1$ we must have $(b^a, a) = 1$ so $b^a \mid d^b$. Similarly since $(c,d) = 1$ we must have $(d^b, c) = 1$ so $d^b \mid b^a$. Therefore $d^b = b^a$.
Suppose $p \mid d$ then $p \mid d^b = b^a \Rightarrow p \mid b$.
Suppose $\nu_p(d) = m, \nu_p(b) = n$ we must have $bm = \nu_p(d^b) = \nu_p(b^a) = an$, ie $b = \frac{an}{m}$.
Note that $p^n \mid b \Rightarrow p^n \mid n \frac{a}{m} \Rightarrow p^n \mid n \Rightarrow p^n \leq n$. Since $(p,a) = 1$..
But since $p^n > n$ if $p \geq 2$ we must have that $b = 1$.
Therefore suppose $r = \frac{a}{b}$ with $(a,b) = 1$ an $r^r = \frac{c}{d}$ we must ahve $a^ac^b = b^ad^b$ and so $b = 1$ implying $r$ is an integer.
\end{questionparts}
Nearly as many attempted this as question 4, but only achieving a quarter of the marks making it the least successfully answered question. Almost all missed the point of the question given in the first sentence, and made other assumptions, which frequently only applied to primes rather than integers in general. As a consequence, most did not satisfactorily justify their results. They generally fared better tackling the second part of (i), though some tried to prove the statement in the wrong direction. They approached (ii) better though few gave a valid argument.