Puzzle challenge

Thursday 12th March 2020

This puzzle was MEI's Maths Item of the Month in August 2011. An archive of puzzles going back to 2006 is available on the MEI websiteOpens a new window. You can also view the puzzles categorisedOpens a new window by GCSE and A level Mathematics topics. The puzzle below is aimed at AS level students, but would also be accessible to bright Year 11 students.

\(\frac{1}{3}+\frac{1}{5}=\frac{8}{15}\;and\;(8,15,17)\) is a Pythagorean triple.

Add the reciprocals of any two consecutive odd numbers. Will the resulting fraction, \(\frac{x}{y}\), always generate an integer Pythagorean triple, \((x,y,z)\)?

By writing the consecutive odd numbers as \(2n-1\) and \(2n+1\) the sum of the fractions is:

\[\frac{1}{2n-1}+\frac{1}{2n+1}=\frac{(2n+1)+(2n-1)}{(2n-1)(2n+1)}=\frac{4n}{4n^2-1}\]

The conjecture is that this generates a fraction \(\frac{a}{b}\) such that \(a^2+b^2=c^2\) where \(a\), \(b\) and \(c\) are all integers.

Squaring and adding the terms for \(a\) and \(b\) gives:

\[(4n)^2+(4n^2-1)^2=16n^2+16n^4-8n^2+1\]

\[=16n^4+8n^2+1\]

\[=(4n^2+1)^2\]

\(4n^2+1\) will be an integer, therefore \((4n,\;4n^2-1,\;4n^2+1)\) is a Pythagorean triple.

By Sheila Eastwood

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