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# Maths Week: Your Thursday puzzle

Fancy another mathematics challenge? (And get the answer to yesterday’s puzzle.)

MATHS WEEK STARTED on Saturday and, as is our annual tradition, we’re setting our readers some puzzles. Give them a go!

Day six - Fermat’s Last Theorem

Yesterday we met Pierre de Fermat (1607 – 1665) in connection with his work on probability. He is probably best known today for Fermat’s Last Theorem, which he had written in the margin of a book and claimed to have a proof too long to fit in the margin. This theorem however has long roots stretching back centuries through many cultures.

There exist triples of whole numbers (which we can call a, b and c) such that a² + b² = c².

These have been found in ancient Babylonian clay tablets almost 4,000 years old and they were studied in ancient India and Greece.

It turns out that if a triangle has sides whose lengths satisfy this relationship, then that triangle must be a right-angled triangle.

In this context, the relationship is known as Pythagoras’ Theorem where a and b are the lengths of two sides of a right-angle triangle and c is then the length of the third (longest) side, the hypotenuse.

The fact that 9+16=25 can be written as 3²+4²=5² has a geometric interpretation.

If a rectangular room measures 3 metres by 4 metres then two diametrically opposite corners are exactly 5 metres apart.

Fermat’s last Theorem states that there are no such triples for powers higher than 2.

That means we can’t find three positive whole numbers a, b, c, such that a³+b³=c³. This was definitively shown in 1770 by “the Mozart of mathematics”, Leonhard Euler (pronounced Oiler).

By 1825 it was shown that there were no solutions for a⁴+b⁴=c⁴ or a⁵+b⁵=c⁵ ? But the search continued for a general proof or for an exception that would disprove the conjecture.

One of the first people to made big progress on this problem was Sophie Germain (1776-1831). The final steps were taken in the 1990s by Andrew Wiles. He had read about this famous unsolved problem when he was a boy in England and was determined to grow up and solve it. This he did, and the proof took him seven years to complete! That proof was 129 pages long and used mathematics that was unknown in Fermat’s time, so while it is not believed he had a proof, if he had it definitely wouldn’t have fit in the margin.

Here are your puzzles for today.

1) A farmer wishes to set out two fences at right angles. She has no surveying equipment or right angles with which to measure. She only has a circular rope 30m long and a measuring tape and of course the fence posts. How can she do it?

2) Can you tell that 8³+13³=14³ cannot be true without multiplying it out in full?

3) Proving Fermat’s Last Theorem took centuries. Disproving it would take only one
counterexample. In one such case Time magazine reported that a mathematician had found a whole number N (which he refused to reveal!) such that 1324ᴺ+731ᴺ=1961ᴺ. A reporter on The New York Times without using any advanced maths proved that this was impossible. How?

Fermat would consider all the possible future possibilities. Remember that the score of the abandoned game is HHT. The possible outcomes are HH, HT, TH and TT.

Only the last of these eventualities will bring Ben out on top. Ben therefore has a 1 in 4 and Aine a 3 in 4 chance. Therefore, Aine would get €7.50 and Ben €2.50.

With three tosses of a coin, there is an even chance of getting a least two tails.

The chances of getting two heads in a row is 3 out of 8.

If two coins are tossed and one is shown to be heads then there is only a 1 in 3 chance that the other coin is also heads. That is because the possible outcomes from tossing two coins are HH, HT, TH, TT. The last of these has no head, so there are three possible conditions: HH,HT,TH and only one of these has two heads.

Come back tomorrow at 7.30pm for the answers to today’s puzzle.

The puzzles this week have been compiled by Eoin Gill and Colm Mulcahy of Maths Week Ireland / South-East Technological University (SETU).

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The Journal Team