1. Activity 1-13: Descent www.carom-maths.co.uk

2. This problem is due to Euler. Task: Show that the equation x 3 + 2y 3 + 4z 3 = 0 has the sole solution (0, 0, 0) for integer x, y and z. Suppose we DO have a solution. We can see that x 3 is even, so x must be even. Say x = 2x’. Thus 8x’ 3 + 2y 3 + 4z 3 = 0, So 4x’ 3 + y 3 + 2z 3 = 0 Thus y = 2y’, and we have 4x’ 3 + 8y’ 3 + 2z 3 = 0, or 2x’ 3 + 4y’ 3 + z 3 = 0. Thus z = 2z’, and we have 2x’ 3 + 4y’ 3 + 8z’ 3 = 0, or x’ 3 + 2y’ 3 + 4z’ 3 = 0.

3. So as long as (x, y, z) is not (0, 0, 0), our solution (x, y, z) has led to a smaller solution (x’, y’, z’). We cannot go on like this indefinitely, since there are finitely many integers between x and 0. So the only possible values for (x, y, z) are (0, 0, 0). Neat? This is an example of the DESCENT method, as used by Pierre de Fermat and many other number theorists.

4. The technique of descent works like this: 2. assume we DO have a solution in integers. 1.to prove some number theoretical statement about integers has no solutions... 3. show that from this solution, we can derive a smaller one, also in integers. 4. we cannot continue this indefinitely, so no solution in integers can exist.

5. Theorem to prove by descent: If gcd(a, b) = 1, and ab = x 2, then a = y 2, b = z 2. Definition: gcd(x, y) = greatest common divisor of x and y Here is another example of this technique at work... So gcd(3, 4) = 1, while gcd(12, 18) = 6. This is sometimes called ‘the highest common factor of x and y’.

6. Suppose a, b provide a counterexample. a = 1  b = x 2, and this is NOT a counterexample, so a > 1. So a has a prime divisor p, and so p|x 2, and so p|x. Thus p 2 |x 2 = ab, and since a and b are coprime, p 2 |a. But that means that a’=a/p 2 and b’ = b provide a smaller counterexample. So by descent, no such counterexample can exist.

7. Theorem (not proved here) If (a, b, c) where a 2 + b 2 = c 2 are the integer sides of a right-angled triangle where gcd(a, b, c) = 1, then (a, b, c) = (m 2 - n 2, 2mn, m 2 + n 2 ) for some integers m and n. The integers m and n will be of opposite parity with gcd(m, n) = 1. Note that a is odd and b is even here. This gives a parametrisation for primitive Pythagorean Triples, that is, where gcd(a, b, c) = 1. For example, if (a, b, c) = (3, 4, 5), then m = 2, n = 1. Fermat’s Last Theorem is extremely tough to prove, BUT the special case for n = 4 can be proved by descent. To help with that, we will use this result:

8. To prove: x 4 + y 4 = z 4 has no solutions for integer x, y, z. In fact, we will prove slightly more than this: x 4 + y 4 = z 2 has no solutions for integer x, y, z. Let’s assume we DO have a solution where x 4 + y 4 = z 2 for integer x, y, z. Then clearly we can insist that gcd(x, y) = gcd(y, z) = gcd (z, x) = gcd (x, y, z) = 1. Fermat’s Last Theorem for n = 4

9. By our parametrisation of primitive Pythagorean triples, we now have x 2 = u 2 – v 2 (which is odd), y 2 = 2uv (which is even) and z = u 2 + v 2 (which is odd). So now we have that v 2 + x 2 = u 2, which gives us x = m 2 – n 2, v = 2mn, u = m 2 + n 2.

10. Now this means that y 2 = 4mnu, and so from our Result 1, m, n and u are all squares. Let’s say that m = p 2, n = q 2 and u = r 2. Thus our equation u = m 2 + n 2 becomes p 4 + q 4 = r 2.

11. And so we have another solution to our starting equation x 4 + y 4 = z 2 using smaller integers than in our original solution. So by descent, x 4 + y 4 = z 2 (and so x 4 + y 4 = z 4 ) has no solutions for integer x, y, z. We are done! We can use the descent method to prove the insolubility of similar equations, like x 4 + 2y 4 = z 4. In fact, x 4 + dy 4 = z 4 for any reasonably small d for which there are no nontrivial solutions will do, (although some are more difficult than others...)

12. With thanks to: Graham Everest, Shaun Stevens and Tom Ward. Carom is written by Jonny Griffiths, hello@jonny-griffiths.nethello@jonny-griffiths.net