Show that for any integers a, b, c, not all zero, and such that |a|, |b|, |c| < 10^{6}, we have |a + b √2 + c √3| > 10^{-21}. But show that we can find such a, b, c with |a + b √2 + c √3| < 10^{-11}.

**Solution**

This needs a trick: (a + b √2 + c √3)(a + b √2 - c √3)(a - b √2 + c √3)(a - b √2 - c √3) = a^{4} + 4b^{4} + 9c^{4} - 4a^{2}b^{2} - 6c^{2}a^{2} - 12b^{2}c^{2} which is an integer. Moreover, it is non-zero by a slight extension of the argument used to prove the irrationality of √2.

For suppose a + b √2 = c √3 (*). Squaring: 2ab √2 = 3c^{2} - a^{2} - 2b^{2}. But √2 is irrational, so a or b is zero. But b cannot be zero since √3 is irrational, so a must be zero. Take b and c relatively prime (by dividing out any common factor if necessary). We have 2b^{2} = 3c^{2}. By the usual argument 2 divides both b and c. Contradiction. Hence (*) is impossible.

So the identity (a + b √2 + c √3)(a + b √2 - c √3)(a - b √2 + c √3)(a - b √2 - c √3) = a^{4} + 4b^{4} + 9c^{4} - 4a^{2}b^{2} - 6c^{2}a^{2} - 12b^{2}c^{2} gives that |a + b √2 + c √3| ≥ 1/|(a + b √2 - c √3)(a - b √2 + c √3)(a - b √2 - c √3)|. But each of the three numbers in this product is at most 10^{7}, so |a + b √2 + c √3| ≥ 10^{-21}.

For the other inequality we use a pigeon-hole argument. Take the 10^{18} numbers a + b √2 + c √3 with 0 ≤ a, b, c < 10^{6}. They all lie in the interval [0, 5 x 10^{6}], so if we divide this interval into 6 x 10^{17} equal parts, then at least two numbers must lie in the same part. Subtracting them gives a number with absolute value at most 5/6 10^{-11}.

© John Scholes

jscholes@kalva.demon.co.uk

10 Dec 1999

Last corrected/updated 21 Nov 02