OCR MEI C3 — Question 13 6 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof
TypePythagorean triples and number patterns
DifficultyStandard +0.3 This is a straightforward algebraic verification question. Part (i) requires expanding $(2t)^2 + (t^2-1)^2$ and showing it equals $(t^2+1)^2$ - routine algebra with no conceptual difficulty. Part (ii) involves solving $20^2 + 21^2 = c^2$ (basic arithmetic) then checking whether this triple fits the given form by testing if any integer $t$ works - a simple verification that requires minimal problem-solving. The question is slightly easier than average as it's mostly mechanical manipulation with clear instructions.
Spec1.01a Proof: structure of mathematical proof and logical steps

Positive integers \(a\), \(b\) and \(c\) are said to form a Pythagorean triple if \(a^2 + b^2 = c^2\).
  1. Given that \(t\) is an integer greater than 1, show that \(2t\), \(t^2 - 1\) and \(t^2 + 1\) form a Pythagorean triple. [3]
  2. The two smallest integers of a Pythagorean triple are 20 and 21. Find the third integer. Use this triple to show that not all Pythagorean triples can be expressed in the form \(2t\), \(t^2 - 1\) and \(t^2 + 1\). [3]

Positive integers $a$, $b$ and $c$ are said to form a Pythagorean triple if $a^2 + b^2 = c^2$.

\begin{enumerate}[label=(\roman*)]
\item Given that $t$ is an integer greater than 1, show that $2t$, $t^2 - 1$ and $t^2 + 1$ form a Pythagorean triple. [3]

\item The two smallest integers of a Pythagorean triple are 20 and 21. Find the third integer.

Use this triple to show that not all Pythagorean triples can be expressed in the form $2t$, $t^2 - 1$ and $t^2 + 1$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C3  Q13 [6]}}