Activity 1-13: Descent

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

2. This problem is due to Euler. Task: Show that the equation x3 + 2y3 + 4z3 = 0 has the sole solution (0, 0, 0) for integer x, y and z. Suppose we DO have a solution. We can see that x3 is even, so x must be even. Say x = 2x’. Thus 8x’3 + 2y3 + 4z3 = 0, So 4x’3 + y3 + 2z3 = 0 Thus y = 2y’, and we have 4x’3 + 8y’3 + 2z3 = 0, or 2x’3 + 4y’3 + z3 = 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 DESCENTmethod, as used by Pierre de Fermat and many other number theorists.

4. The technique of descent works like this: • to prove some number theoretical statement • about integers has no solutions... 2. assume we DO have a solution in integers. 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. Here is another example of this technique at work... Definition: gcd(x, y) = greatest common divisor of x and y This is sometimes called ‘the highest common factor of x and y’. So gcd(3, 4) = 1, while gcd(12, 18) = 6. Theorem to prove by descent: If gcd(a, b) = 1, and ab = x2, then a = y2, b = z2.

6. Suppose a, b provide a counterexample. a = 1  b = x2, and this is NOT a counterexample, so a > 1. So a has a prime divisor p, and so p|x2, and so p|x. Thus p2|x2 = ab, and since a and b are coprime, p2|a. But that means that a’=a/p2and b’ = b provide a smaller counterexample. So by descent, no such counterexample can exist.

7. 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: Theorem (not proved here) If (a, b, c) where a2 + b2 = c2 are the integer sides of a right-angled triangle where gcd(a, b, c) = 1, then (a, b, c) = (m2 - n2, 2mn, m2 + n2) 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. For example, if (a, b, c) = (3, 4, 5), then m = 2, n = 1. This gives a parametrisation for primitive Pythagorean Triples, that is, where gcd(a, b, c) = 1.

8. Fermat’s Last Theorem for n = 4 To prove: x4 + y4 = z4has no solutions for integerx, y, z. In fact, we will prove slightly more than this: x4 + y4 = z2has no solutions for integer x, y, z. Let’s assume we DO have a solution where x4 + y4 = z2 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.

9. By our parametrisation of primitive Pythagorean triples, we now have x2 = u2 – v2(which is odd), y2 = 2uv (which is even) and z = u2 + v2 (which is odd). So now we have that v2 + x2 = u2, which gives us x = m2 – n2, v = 2mn, u = m2 + n2.

10. Now this means that y2 = 4mnu, and so from our Result 1, m, n and u are all squares. Let’s say that m = p2, n = q2and u = r2. Thus our equation u = m2 + n2becomes p4 + q4 = r2.

11. And so we have another solution to our starting equation x4 + y4 = z2 using smaller integers than in our original solution. So by descent, x4 + y4 = z2(and so x4 + y4 = z4) 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 x4 + 2y4 = z4. In fact, x4 + dy4 = z4for 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.net