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.