For which $k$ do there exist $k$ pairwise distinct primes $p_{1}, p_{2}, \ldots, p_{k}$ such that

$$
p_{1}^{2}+p_{2}^{2}+\cdots+p_{k}^{2}=2010 ?
$$
We show that it is possible only if $k=7$.

The 15 smallest prime squares are:

$$
4,9,25,49,121,169,289,361,529,841,961,1369,1681,1849,2209
$$

Since $2209>2010$ we see that $k \leq 14$.

Now we note that $p^{2} \equiv 1 \bmod 8$ if $p$ is an odd prime. We also have that $2010 \equiv 2 \bmod 8$. If all the primes are odd, then writing the original equation modulo 8 we get

$$
k \cdot 1 \equiv 2 \bmod 8
$$

so either $k=2$ or $k=10$.

$k=2:$ As $2010 \equiv 0 \bmod 3$ and $x^{2} \equiv 0$ or $x^{2} \equiv 1 \bmod 3$ we conclude that $p_{1} \equiv p_{2} \equiv 0$ mod 3. But that is impossible.

$k=10$ : The sum of first 10 odd prime squares is already greater than $2010(961+841+$ $529+\cdots>2010)$ so this is impossible.

Now we consider the case when one of the primes is 2. Then the original equation modulo 8 takes the form

$$
4+(k-1) \cdot 1 \equiv 2 \quad \bmod 8
$$

so $k \equiv 7 \bmod 8$ and therefore $k=7$.

For $k=7$ there are 4 possible solutions:

$$
\begin{aligned}
4+9+49+169+289+529+961 & =2010 \\
4+9+25+121+361+529+961 & =2010 \\
4+9+25+49+121+841+961 & =2010 \\
4+9+49+121+169+289+1369 & =2010
\end{aligned}
$$

Finding them should not be too hard. We are already assuming that 4 is included. Considerations modulo 3 show that 9 must also be included. The square 1681 together with the 6 smallest prime squares gives a sum already greater than 2010, so only prime squares up to $37^{2}=1369$ can be considered. If 25 is included, then for the remaining 4 prime squares considerations modulo 10 one can see that 3 out of 4 prime squares from $\{121,361,841,961\}$ have to be used and two of four cases are successful. If 25 is not included, then for the remaining 5 places again from considerations modulo 10 one can see, that 4 of them will be from the set $\{49,169,289,529,1369\}$ and two out of five cases are successful.