Problems nobody has solved
Tonight's idea
Not every question in mathematics has an answer yet. Here are four that do not: easy to state, easy to test, unproved.
Collatz. Take a whole number. If it is even, halve it. If odd, triple it and add 1. Repeat. The claim: you always reach 1. Every start checked by computer, beyond \(2^{68}\), does. Nobody can prove all do.
Goldbach. Every even number above 2 is a sum of two primes: \(8 = 3 + 5\), \(20 = 7 + 13\). Goldbach proposed it in a letter to Euler. Checked up to \(4 \times 10^{18}\). No proof.
Twin primes. Primes two apart, like 11 and 13. Do the pairs go on forever? Nobody knows. It is proved that infinitely many pairs of primes are at most 246 apart. Not 2, yet.
Odd perfect numbers. A number is perfect if its proper divisors (all divisors except itself) add up to it: \(6 = 1 + 2 + 3\). Every known perfect number is even. An odd one, if it exists, has over 1500 digits.
Why so hard? A billion checked cases prove nothing about the next. A proof must cover every number at once.
Watch one: Collatz from 27
Small starts finish fast: \(6 \to 3 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1\), eight steps. Then try 27:
\[27 \to 82 \to 41 \to 124 \to 62 \to 31 \to 94 \to 47 \to 142 \to 71 \to 214 \to \cdots\]
It climbs to 9232 before falling, and takes 111 steps to reach 1. No start below 27 takes anywhere near as long. That is the trouble: a rule this simple produces behaviour nobody can predict without running it.
Try any start and watch the flight: mathdetours.com/demos/collatz.html.
Your turn
- Run the Collatz rule from 7 and from 11. How many steps does each take to reach 1, and what is the highest number each one visits?
- Write 36 as a sum of two primes in every possible way (count \(5 + 31\) and \(31 + 5\) once). Then do the same for 98.
- (a) List every twin prime pair below 100. There are eight. (b) Look at the number sitting between each pair: 4, 6, 12, 18, … Explain why, from the pair (5, 7) onwards, that middle number is always a multiple of 6.
- (a) Show that 496 is perfect: list its divisors (they are \(1, 2, 4, 8, 16\) and \(31, 62, 124, 248, 496\)) and add the proper ones. (b) Show that a power of 2 is never perfect: add the proper divisors of \(2^k\). How far short does the sum fall?
Stretch: if Goldbach is right, then…
Suppose Goldbach's claim is true for every even number. Show that every odd number greater than 5 is then a sum of three primes. Then explain why the Collatz claim only needs proving for odd starting numbers.
Answers
- From 7: \(7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1\), 16 steps, peak 52. From 11 the sequence joins the same path at 11, so 14 steps, peak 52.
- \(36 = 5 + 31 = 7 + 29 = 13 + 23 = 17 + 19\), four ways. \(98 = 19 + 79 = 31 + 67 = 37 + 61\), three ways.
- (a) (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73). (b) Call the pair \(p\) and \(p + 2\). Both are odd, so \(p + 1\) is even. Of the three consecutive numbers \(p, p+1, p+2\), exactly one is a multiple of 3. It cannot be \(p\) or \(p + 2\), since those are primes bigger than 3. So \(p + 1\) is a multiple of 2 and of 3, hence of 6.
- (a) \(1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496\). (b) The proper divisors of \(2^k\) are \(1, 2, 4, \dots, 2^{k-1}\), and they add to \(2^k - 1\). (Check: \(1 + 2 + 4 + 8 = 15 = 16 - 1\).) The sum is always exactly 1 short, so no power of 2 is perfect.
Stretch. Let \(n\) be odd and greater than 5. Then \(n - 3\) is even and at least 4, so by Goldbach \(n - 3 = p + q\) for primes \(p, q\), and \(n = 3 + p + q\). For Collatz: an even start is halved, and halved again, until it becomes odd. If every odd number reaches 1, every even number reaches an odd number and then reaches 1.