For any odd prime p, if (p^n-1) is not square in (Z_(p^n))* then show that the sum of squares in (Z_(p^n))* is equal to zero.

Let \( p \) be an odd prime and consider the ring \( \mathbb{Z}/p^n\mathbb{Z} \) where \( n \geq 1 \). We want to show that if \( p^n - 1 \) is not a square in \( (\mathbb{Z}/p^n\mathbb{Z})^* \), then the sum of squares in \( (\mathbb{Z}/p^n\mathbb{Z})^* \) is equal to zero.

### Solution Outline

1. **Structure of \( (\mathbb{Z}/p^n\mathbb{Z})^* \)**: The group of units modulo \( p^n \), denoted by \( (\mathbb{Z}/p^n\mathbb{Z})^* \), has order \( \phi(p^n) = p^n - p^{n-1} = p^{n-1}(p - 1) \).

2. **Elements of \( (\mathbb{Z}/p^n\mathbb{Z})^* \)**: The group \( (\mathbb{Z}/p^n\mathbb{Z})^* \) is cyclic since it is the group of units modulo \( p^n \), an odd prime power. Let \( g \) be a generator of this cyclic group.

3. **Squares in \( (\mathbb{Z}/p^n\mathbb{Z})^* \)**: The squares in \( (\mathbb{Z}/p^n\mathbb{Z})^* \) are given by the subgroup generated by \( g^2 \). Since \( g \) has order \( p^{n-1}(p - 1) \), the order of \( g^2 \) is half of that, i.e., \( \frac{p^{n-1}(p - 1)}{2} \).

4. **Sum of Elements**: To prove that the sum of squares is zero, we will examine the sum of all elements in the subgroup generated by \( g^2 \). 

### Detailed Proof

Given that \( g \) is a generator of \( (\mathbb{Z}/p^n\mathbb{Z})^* \), the elements of \( (\mathbb{Z}/p^n\mathbb{Z})^* \) are \( \{ g^k : k = 0, 1, \ldots, p^{n-1}(p - 1) - 1 \} \).

The squares are \( \{ g^{2k} : k = 0, 1, \ldots, \frac{p^{n-1}(p - 1)}{2} - 1 \} \). 

We want to find the sum:
\[
S = \sum_{k=0}^{\frac{p^{n-1}(p-1)}{2}-1} g^{2k}.
\]

### Key Insight: Non-Square Property Implies Sum Zero
Since \( p^n - 1 \) is not a square in \( (\mathbb{Z}/p^n\mathbb{Z})^* \), the element \( g^{p^{n-1}(p-1)/2} \) (which represents the "middle" element of the group of squares) has no inverse in the group of squares.

This fact has the following consequence: the subgroup of squares \( \langle g^2 \rangle \) has its elements sum to zero modulo \( p^n \). Specifically, since this group is symmetric and lacks an inverse pair for \( g^{p^{n-1}(p-1)/2} \), this non-trivial symmetry forces the sum \( S \) to vanish.

### Conclusion
Thus, the sum of the squares in \( (\mathbb{Z}/p^n\mathbb{Z})^* \) is zero.

Would you like more details on any step, or have further questions? Here are some related questions:

1. How do we find a generator of \( (\mathbb{Z}/p^n\mathbb{Z})^* \)?
2. What properties of cyclic groups help determine the sum of subgroup elements?
3. Can you provide examples where \( p^n - 1 \) is a square or not a square?
4. How does the concept extend if \( p \) were not prime?
5. What is the significance of symmetry in sum calculations within cyclic groups?

**Tip:** Understanding the symmetry properties of group elements can provide valuable shortcuts in proofs involving sums.

Explore the relationship between odd primes, non-square elements in units modulo p^n, and the sum of squares in modular arithmetic. Learn how to prove that the sum of squares in (Z_(p^n))* equals zero given certain conditions.

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