## Why 17?

In Plato's Theaetetus we learn that mathematician Theodorus has established irrationality of square roots for all non-square numbers up to 17 and then he stopped. This is definitely a curious question as to what happened, why he stopped at 17. There is no documentation that would point to a method of proof he used. We may only guess and speculate.

In a 1976 paper, McCabe offers this reasoning for a possible Pythagorean proof:

 The Pythagorean proof of the irrationality of √2 is well known. If we assume that √d = a/b and that a/b is in lowest terms, then 2b² = a² implies that a² and hence that a is even, that is, a = 2k. Substituting 2k for a yields 2b² = 4k² ,or b² = 2k². Therefore b² and hence b is even. But a and b cannot both be even if a/b is in lowest terms.

A slightly different argument is offered in the collection of other irrationality proofs of 2.

It is noteworthy that neither appeals to the general properties of prime numbers which were probably unknown at the time of Pythagoras. Instead of claiming that 2 is a prime and invoking the Fundamental Theorem of Arithmetic, the above argument plays on the juxtaposition of odd and even numbers. It directly applies to non-square even numbers, while the following might be considered the application of that idea to proving the irrationality of 3.

Assume 3 = a/b, in lowest terms. Then a² = 3b². If a is even then so is 3b² and hence b. Similarly, if b is even, so is a. Therefore, the two are odd. Assume a = 2k + 1 and b = 2m + 1. Then after an easy algebra

 2 = 4(k² + k - 3(m² + m)),

or

 1 = 2(k² + k - 3(m² + m)),

which is impossible since the left-hand side is odd, whereas the right-hand side is even.

This argument adapts to other odd numbers with slight modifications. Easily to the numbers, 7, 11, 15, ... in the form 4n + 3, and with more elaboration for the numbers 5 and 13. For example, assuming 5 = a/b, we are lead to

 4 = 4(k² + k - 5(m² + m)),

and then to

 1 = k² + k - 5(m² + m),

where we observe that the sum of a number with its square is always even, making even the right-hand side. The argument works for 13 but becomes cumbersome for proving the irrationality of 17.

McCabe suggests that a theorem by W. R. Knorr may have been the foundation for the Pythagorean reasoning:

### Theorem

If p is a positive integer which can be written in any one of the following forms, 4n + 2, 4n + 3, or 8n + 5, for n = 0, 1, 2, ..., then p can be proved irrational using only even-odd techniques.

Assuming, for example that 8n + 5 = a/b, in lowest terms, we eventually arrive at

 8nm² + 8nm + 2n + 5(m² + m) + 1 = k² + k.

with an odd number on the left and even number on the right.

The only odd numbers not covered by the theorem are those in the form 8n + 1: 1, 9, 17, ... 9 is itself a square, so the difficulty arises first for 17.

Hardy and Wright up to the 1960 edition of their classical book tried to extend the odd/even argument in a different way. For example, for 5, they observed that every integer is in one of the following forms: 5n, 5n + 1, 5n + 2, 5n + 3, 5n + 4. Squaring shows that only (5n)² is divisible by 5. So that 5 = a/b would imply that a is divisible by 5 and then so is b. A contradiction. However, this argument applies with relative ease to 17 and beyond. Up to the 1960 edition, Hardy and Wright were countering this fact by the possibility that Theodorus might have got tired applying this approach just before reaching 17. In later editions, the book has adopted McCabe's reasoning.

Below is another attempt along the same lines by a notable Russian historian of mathematics, to reverse engineer the Pythagorean thought process.

Assume more generally that we'd like to demonstrate irrationality of d, for some d, a non-square whole number. If d is even we can proceed as in the case d = 2. So assume d is odd and d = a/b (in lowest terms); then

 (1) a2 = db2.

As before, a and b must be both odd. Assuming a = 2k + 1 and b = 2m + 1 and substituting into (1) we get after regrouping

 8(k(k + 1)/2 - dm(m + 1)/2) = d - 1.

The left hand side here is divisible by 8 and so the right hand side must be of the form 8t, so that we ought to have d = 8t + 1. If d is not like that, we'd arrive at a contradiction as when d is one of 3, 5, 7, 11, 13, and 15. But if d = 8t + 1, no contradiction arises and the proof fails. This happens for 9 (but 9 is a complete square) and 17. This is why Theodorus stopped at this number.

### References

1. I. G. Bashmakova, A. I. Lapin, Pifagor, Kvant, no 1, 1986, p. 10 (in Russian)
2. G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth Edition, Oxford Science Publications, 1996.
3. R. L. McCabe, Theodorus' Irrationality Proofs, Math Magazine, Vol. 49, No. 4. (Sep., 1976), pp. 201-203. 