4. Find the smallest integer $n \geqslant 1$ such that the equation

$$
a^{2}+b^{2}+c^{2}-n d^{2}=0
$$

admits the only integer solution $a=b=c=d=0$.

## Solutions
Solution 1 First, we check that no integer $n \leqslant 6$ works. Indeed, the quintuplets $(a, b, c, d, n)$ equal to

$$
(1,0,0,1,1),(1,1,0,1,2),(1,1,1,1,3),(2,0,0,1,4),(2,1,0,1,5) \text { and }(2,1,1,1,6)
$$

are solutions to the equation.

We will now show that $n=7$ is the integer we are looking for. Indeed, let $(a, b, c, d)$ be a solution to the equation in the statement, such that $|d|$ is minimal.

If $d$ is odd,

$$
0 \equiv a^{2}+b^{2}+c^{2}-n d^{2} \equiv a^{2}+b^{2}+c^{2}+1 \quad(\bmod 8)
$$

However, the only squares modulo 8 are 0, 1, and 4. The above relation is therefore impossible.

Thus, $d$ is even and

$$
0 \equiv a^{2}+b^{2}+c^{2}-n d^{2} \equiv a^{2}+b^{2}+c^{2} \quad(\bmod 4)
$$

However, the only squares modulo 4 are 0 and 1. Consequently, $a^{2} \equiv b^{2} \equiv c^{2} \equiv 0(\bmod 4)$, so $a, b$, and $c$ are all even.

In conclusion, $(a / 2, b / 2, c / 2, d / 2)$ is also a solution to the problem. Since $|d| / 2 \geqslant|d|$, we conclude that $d=0$, and therefore $a=b=c=0$ as well. Thus, $n=7$ is indeed the integer we are looking for.