A prime number $p$ is a [b]moderate[/b] number if for every $2$ positive integers $k > 1$ and $m$, there exists k positive integers $n_1, n_2, ..., n_k $ such that \[ n_1^2+n_2^2+ ... +n_k^2=p^{k+m} \]
If $q$ is the smallest [b]moderate[/b] number, then determine the smallest prime $r$ which is not moderate and $q < r$.
1. **Identify the smallest moderate prime number \( q \):**
   - A prime number \( p \) is considered moderate if for every \( k > 1 \) and \( m \), there exist \( k \) positive integers \( n_1, n_2, \ldots, n_k \) such that:
     \[
     n_1^2 + n_2^2 + \cdots + n_k^2 = p^{k+m}
     \]
   - We need to determine the smallest prime \( q \) that satisfies this condition.

2. **Check if \( p = 2 \) is moderate:**
   - For \( k = 2 \) and \( m = 2 \), we need to check if \( 2^{2+2} = 16 \) can be expressed as the sum of two non-zero squares.
   - The possible pairs \((a, b)\) such that \( a^2 + b^2 = 16 \) are \((4, 0)\), \((0, 4)\), \((2, 4)\), and \((4, 2)\). However, none of these pairs consist of two non-zero integers.
   - Therefore, \( p = 2 \) is not moderate.

3. **Consider primes \( p > 2 \):**
   - For \( k = 2 \) and \( m = 1 \), we need \( p^3 = a^2 + b^2 \).
   - If \( p \) divides both \( a \) and \( b \), then \( p = (a/p)^2 + (b/p)^2 \), which implies \( p \equiv 1 \pmod{4} \).
   - If \( p \nmid b \), then \((ab^{-1})^2 \equiv -1 \pmod{p}\), which also implies \( p \equiv 1 \pmod{4} \).
   - Therefore, primes of the form \( p \equiv 3 \pmod{4} \) are not moderate.

4. **Verify \( p \equiv 1 \pmod{4} \) primes:**
   - It is known that primes \( p \equiv 1 \pmod{4} \) can be represented as \( p = a^2 + b^2 \).
   - Products of sums of two squares are also sums of two squares:
     \[
     (x^2 + y^2)(z^2 + t^2) = (xz \pm yt)^2 + (xt \mp yz)^2
     \]
   - For \( p^2 = (a^2 + b^2)^2 = (a^2 - b^2)^2 + (2ab)^2 = A^2 + B^2 \).

5. **Lemma: For any \( k \geq 1 \), \( p^k \) can be represented as a sum of \( k \) non-zero squares:**
   - Base cases:
     - \( p = a^2 + b^2 \)
     - \( p^2 = A^2 + B^2 \)
   - Induction step:
     - For \( k > 2 \), assume \( p^{2\ell + 1} = (a^2 + b^2)p^{2\ell} = (ap^{\ell})^2 + \sum_{j=1}^{2\ell} (ba_j)^2 \), a sum of \( 2\ell + 1 \) non-zero squares.
     - For \( p^{2\ell + 2} = p^2 \cdot p^{2\ell} = p^2 \sum_{j=1}^{2\ell} a_j^2 = \sum_{j=1}^{2\ell-2} (pa_j)^2 + (Aa_{2\ell-1})^2 + (Aa_{2\ell})^2 + (Ba_{2\ell-1})^2 + (Ba_{2\ell})^2 \), a sum of \( 2\ell + 2 \) non-zero squares.

6. **Conclusion:**
   - The smallest moderate prime \( q \) is \( 5 \) since \( 5 \equiv 1 \pmod{4} \).
   - The smallest prime \( r \) which is not moderate and \( q < r \) is \( 7 \) since \( 7 \equiv 3 \pmod{4} \).

The final answer is \( \boxed{ 7 } \).