a, b, c are integers such that a ≥ 0, b ≥ 0, ab ≥ c^{2}. Show that for some n we can find integers x_{1}, x_{2}, ... , x_{n}, y_{1}, y_{2}, ... , y_{n} such that x_{1}^{2} + x_{2}^{2} + ... + x_{n}^{2} = a, y_{1}^{2} + y_{2}^{2} + ... + y_{n}^{2} = b, x_{1}y_{1} + x_{2}y_{2} + ... + x_{n}y_{n} = c.

**Solution**

If x_{i}, y_{i} is a solution for a, b, c, then -x_{i}, y_{i} is a solution of a, b, -c, so it is sufficient to consider non-negative c. Without loss of generality we may assume a ≥ b. If c ≤ b, then it is easy to give an explicit solution: take x_{i} = y_{i} = 1 for c values of i; x_{i} = 1, y_{i} = 0 for a-c values of i; x_{i} = 0, y_{i} = 0 for b-c values of i. However, there is no obvious explicit solution for c > b (and hence a > b > c). So we use induction on a+b.

If a + b = 0, then a = b = c = 0, so we can take n = 1, x_{1} = y_{1} = 0. Suppose it is true for a + b < m. Take a, b, c with a + b = m. As noted above, we may assume a > c > b. Hence a' = a+b-2c, b' = b, c' = c-b is a valid triple with a'+b' < m. So there is a solution x_{i}', y_{i}' for a', b', c'. But now x_{i} = x_{i}' + y_{i}', y_{i} = y_{i}' is a solution for a, b, c.

© John Scholes

jscholes@kalva.demon.co.uk

4 Sep 2002