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A numeric set \(M\) containing 2003 distinct numbers is such that for every two distinct elements \(a, b\) in \(M\), the number \(a^2+ b\sqrt 2\) is rational. Prove that for any \(a\) in \(M\) the number \(q\sqrt 2\) is rational.

Prove the following formulae are true: \[\begin{aligned} a^{n + 1} - b^{n + 1} &= (a - b) (a^n + a^{n-1}b + \dots + b^n);\\ a^{2n + 1} + b^{2n + 1} &= (a + b) (a^{2n} - a^{2n-1}b + a^{2n-2}b^2 - \dots + b^{2n}). \end{aligned}\]