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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? ### Exhaustion

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

### Why do this problem?

The problem provides purposeful practice on inequalities and leads to a proof using mathematical induction.
It also builds on the problems Farey Sequences and Mediant Madness, and provides a foundation for the beautiful and surprising result in Ford Circles.

### Possible approach

For a geometrical approach to proving that $\frac{b}{d} < \frac{a+b}{c+d} < \frac{a}{c}$, you may wish to start with Mediant Madness

To prove algebraically that $\frac{a+b}{c+d} < \frac{a}{c}$, given that $\frac bd < \frac ac$, you may need to offer the hint to rearrange to get $bc < ad$, and add $ac$ to both sides of the inequality.

Next, invite students to use Mediants to quickly calculate the first few Farey Sequences, and calculate $ad-bc$ for a few pairs of Farey Neighbours.

Once they establish that $ad-bc=1$ for the examples they try, invite them to construct a proof by induction to show that it holds for all Farey Neighbours.

### Key questions

What are we trying to show?
Is there anything we can do to both sides of the inequality we have, to get us to the inequality we want?
If two fractions are Farey Neighbours in $F_n$, will they still be Farey Neighbours in $F_{n+1}$?
If they are not Farey Neighbours in $F_{n+1}$, what will the new fraction between them be?

### Possible extension

After working on this problem, students could explore Ford Circles.

### Possible support

Farey Sequences and Mediant Madness can be used to ease students into the exploration of the ideas in this problem.