, 2012

One solution is straightforward: $a = b = c = d = 0$. We will prove that it is the only one. Suppose there is another solution $(a, b, c, d)$. We may suppose that the integers $a, b, c$ and $d$ are coprime, otherwise their greatest common divisor could be deleted from the equation (because not all numbers are equal to 0, their greatest common divisor exists). The equation can be reorganized into $a\sqrt{2} + b\sqrt{5} = d\sqrt{10} - c$ and squared to obtain
$$2a^2 + 5b^2 - c^2 - 10d^2 = 2\sqrt{10}(cd - ab).$$
Because $\sqrt{10}$ is not a rational number, we conclude $2a^2 + 5b^2 - c^2 - 10d^2 = 0$ or
$$2a^2 - c^2 = 10d^2 - 5b^2.$$
Because the right side of the equation is divisible by 5, the same must hold for the left side of the equation. The square of a natural number gives a remainder of 0, 1 or 4 when divided by 5. From this we conclude that the numbers $a^2$ and $c^2$ must give a remainder of 0 when divided by 5, hence $a$ and $c$ must be divisible by 5. The left side of the equation is thus divisible by 25, and the same must hold for the right side of the equation. Consequently, $2d^2 - b^2$ must be divisible by 5. Like before we conclude that $b$ and $d$ are divisible by 5. All numbers $a, b, c$ and $d$ are thus divisible by 5, which is a contradiction since they are coprime.