### 41st Putnam 1980

Problem A4

Show that for any integers a, b, c, not all zero, and such that |a|, |b|, |c| < 106, 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) = a4 + 4b4 + 9c4 - 4a2b2 - 6c2a2 - 12b2c2 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 = 3c2 - a2 - 2b2. 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 2b2 = 3c2. 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) = a4 + 4b4 + 9c4 - 4a2b2 - 6c2a2 - 12b2c2 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 107, so |a + b √2 + c √3| ≥ 10-21.

For the other inequality we use a pigeon-hole argument. Take the 1018 numbers a + b √2 + c √3 with 0 ≤ a, b, c < 106. They all lie in the interval [0, 5 x 106], so if we divide this interval into 6 x 1017 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.