Problem 1227

To find the smallest integer $n \geq 1$ such that the equation $a^2 + b^2 + c^2 - nd^2 = 0$ has $(0,0,0,0)$ as the unique solution, we need to ensure that for any non-zero integers $a$, $b$, $c$, and $d$, the equation does not hold.

1. **Check for $n = 1$:**
$$a^2 + b^2 + c^2 - d^2 = 0$$
For $a = b = 2$, $c = 1$, and $d = 3$:
$$2^2 + 2^2 + 1^2 - 3^2 = 4 + 4 + 1 - 9 = 9 - 9 = 0$$
This equation holds for non-zero integers, so $n = 1$ is not the solution.

2. **Check for $n = 2$:**
$$a^2 + b^2 + c^2 - 2d^2 = 0$$
For $a = b = 1$, $c = 4$, and $d = 3$:
$$1^2 + 1^2 + 4^2 - 2 \cdot 3^2 = 1 + 1 + 16 - 18 = 18 - 18 = 0$$
This equation holds for non-zero integers, so $n = 2$ is not the solution.

3. **Check for $n = 3$:**
$$a^2 + b^2 + c^2 - 3d^2 = 0$$
For $a = b = c = d = 1$:
$$1^2 + 1^2 + 1^2 - 3 \cdot 1^2 = 1 + 1 + 1 - 3 = 3 - 3 = 0$$
This equation holds for non-zero integers, so $n = 3$ is not the solution.

4. **Check for $n = 4$:**
$$a^2 + b^2 + c^2 - 4d^2 = 0$$
We need to show that the only solution is $a = b = c = d = 0$. Suppose there exist non-zero integers $a$, $b$, $c$, and $d$ such that:
$$a^2 + b^2 + c^2 = 4d^2$$
By Fermat's theorem on sums of two squares, a number of the form $4k + 3$ cannot be expressed as the sum of three squares. Since $4d^2$ is always of the form $4k$, it cannot be expressed as the sum of three non-zero squares. Therefore, the only solution is $a = b = c = d = 0$.

Thus, the smallest integer $n$ such that the equation $a^2 + b^2 + c^2 - nd^2 = 0$ has $(0,0,0,0)$ as the unique solution is $n = 4$.

The final answer is $\boxed{ n = 4 }$.