OCR H240/02 2020 November — Question 5 5 marks

Exam BoardOCR
ModuleH240/02 (Pure Mathematics and Statistics)
Year2020
SessionNovember
Marks5
PaperDownload PDF ↗
TopicProof by induction
TypeProve polynomial divisibility property
DifficultyStandard +0.3 This is a straightforward number theory problem requiring basic algebraic manipulation. Part (a) needs recognizing that both expressions are integers when n is odd (simple parity argument). Part (b) is routine verification by substitution into the Pythagorean theorem formula. The question is slightly easier than average as it guides students through the steps and requires only standard algebraic techniques without novel insight.
Spec1.01a Proof: structure of mathematical proof and logical steps

  1. Determine the set of values of \(n\) for which \(\frac{n^2 - 1}{2}\) and \(\frac{n^2 + 1}{2}\) are positive integers. [3]
A 'Pythagorean triple' is a set of three positive integers \(a\), \(b\) and \(c\) such that \(a^2 + b^2 = c^2\).
  1. Prove that, for the set of values of \(n\) found in part (a), the numbers \(n\), \(\frac{n^2 - 1}{2}\) and \(\frac{n^2 + 1}{2}\) form a Pythagorean triple. [2]

\begin{enumerate}[label=(\alph*)]
\item Determine the set of values of $n$ for which $\frac{n^2 - 1}{2}$ and $\frac{n^2 + 1}{2}$ are positive integers. [3]
\end{enumerate}

A 'Pythagorean triple' is a set of three positive integers $a$, $b$ and $c$ such that $a^2 + b^2 = c^2$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Prove that, for the set of values of $n$ found in part (a), the numbers $n$, $\frac{n^2 - 1}{2}$ and $\frac{n^2 + 1}{2}$ form a Pythagorean triple. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR H240/02 2020 Q5 [5]}}