3. Determine all representations of the number 2001 as a sum of 1979 squares of natural numbers.
Solution. Let $x_{1}, x_{2}, \ldots, x_{1979}$ be natural numbers for which

$$
x_{1} \geq x_{2} \geq \ldots \geq x_{1979} \text { and } x_{1}^{2}+x_{2}^{2}+\ldots+x_{1979}^{2}=2001
$$

It is not possible for all these numbers to be greater than 1, because then we would have

$$
x_{1}^{2}+x_{2}^{2}+\ldots+x_{1979}^{2} \geq 1979 \cdot 2^{2}>2001
$$

Let $x_{1} \geq x_{2} \geq \ldots \geq x_{k}>1=x_{k+1}=\ldots=x_{1979}$. Then we have

$$
k x_{k}^{2} \leq x_{1}^{2}+x_{2}^{2}+\ldots+x_{k}^{2}=2001-1979+k=k+22
$$

which means $4 \leq x_{k}^{2} \leq 1+\frac{22}{k}$, from which we get $k \leq 7$. Since $x_{1}^{2}+x_{2}^{2}+\ldots+x_{7}^{2}=29$, we have $x_{1}, x_{2}, \ldots, x_{7} \in\{1,2,3,4\}$, (Otherwise, $x_{1}^{2}+x_{2}^{2}+\ldots+x_{7}^{2} \geq 5^{2}+6=31$.) Let among the numbers $x_{1}, x_{2}, \ldots, x_{7}$ there be $a, b, c, d$ ones, twos, threes, fours, respectively. Then

$$
a+b+c+d=7, a+4 b+9 c+16 d=29
$$

so we have

$$
22=a+4 b+9 c+16 d-(a+b+c+d)=3 b+8 c+15 d
$$

$c \leq 2$ and $3 \mid 22-8 c$. This is only possible for $c=2$. In this case, $15 d+3 b=6$, so $d=0$ and $b=2$. Thus, $a=7-(b+c+d)=3$. It is easily verified that

$$
2001=2 \cdot 3^{2}+2 \cdot 2^{2}+1975 \cdot 1^{2}
$$

which means there is only one decomposition of the number 2001 that satisfies the given conditions.