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Shades of Fermat's Last Theorem

The familiar Pythagorean 3-4-5 triple gives one solution to (x-1)^n + x^n = (x+1)^n so what about other solutions for x an integer and n= 2, 3, 4 or 5?


Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2

Code to Zero

Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135.

Farey Neighbours

Age 16 to 18
Challenge Level
You may wish to explore Farey Sequences and Mediant Madness before working on this problem.

The Farey sequence $F_n$ is the list written in increasing order of all the rational numbers between $0$ and $1$ that have only the numbers $1, 2, 3, ... n$ as denominators. We have $$\eqalign{F_1&=\frac{0}{1}, \frac{1}{1}\cr F_2&=\frac{0}{1}, \frac{1}{2}, \frac{1}{1}.}$$

For the two rational numbers $\frac{a}{c}$ and $\frac{b}{d}$ the mediant is defined as $\frac{a+b}{c+d}$.

Show that if $0< \frac{b}{d} < \frac{a}{c}< 1$ then $\frac{b}{d} < \frac{a+b}{c+d} < \frac{a}{c}$.

Each Farey sequence $F_{n+1}$ must contain all of the terms of $F_{n}$, along with some new terms. Each 'new' term in the Farey sequence $F_{n+1}$ is the mediant of two consecutive terms in $F_n$, but not all mediants of consecutive terms of $F_n$ are included: where the denominator of the mediant is greater than $n+1$ the mediant does not occur in $F_{n+1}$ and instead two consecutive terms in $F_n$ are repeated in $F_{n+1}$.

Use the mediants to work out $F_3, F_4$ and $F_5$.

Two consecutive terms in a Farey sequence are known as Farey Neighbours.

Let $\frac{b}{d}$ and $\frac{a}{c}$ be Farey Neighbours. Choose some pairs of Farey Neighbours from the Farey Sequences you have worked out, and calculate $ad-bc$.
What do you notice?

Can you prove it using mathematical induction?

This is the key result that links Farey sequences to infinite sets of circles that are packed together, each circle touching its neighbours. See the problem Ford Circles.