Verify that ## A(225)\geq 1 ##

[Math100]

Homework Statement

For each real-valued nonprincipal character $\chi\pmod {16}$, verify that $A(225)\geq 1$.

Relevant Equations

Let $\chi$ be any real-valued character mod $k$ and let $A(n)=\sum_{d\mid n}\chi(d)$. Then $A(n)\geq 0$ for all $n$, and $A(n)\geq 1$ if $n$ is a square.

Let $n=225$ and $d$ be the divisors of $n$.
Then $d=\left \{ 1, 3, 5, 9, 15, 25, 45, 75, 225 \right \}$.
Note that the real-valued nonprincipal characters $\chi\pmod {16}$ are $\chi(1), \chi(7), \chi(9), \chi(15)$.
Observe that $\chi(1)=1, \chi(7)=\pm 1, \chi(9)=\pm 1, \chi(15)=\pm 1$.
Thus $A(225)=\sum_{d\mid 225}\chi(d)=\chi(1)+\chi(9)+\chi(15)=1+1+1=3\geq 1$.
Therefore, $A(225)\geq 1$.

\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline n & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\
\hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
\hline \chi_{2}(n) & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \\
\hline \chi_{3}(n) & 1 & i & i & 1 & -1 & -i & -i & -1 \\
\hline \chi_{4}(n) & 1 & -i & i & -1 & -1 & i & -i & 1 \\
\hline \chi_{5}(n) & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 \\
\hline \chi_{6}(n) & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 \\
\hline \chi_{7}(n) & 1 & i & -i & -1 & -1 & -i & i & 1 \\
\hline \chi_{8}(n) & 1 & -i & -i & 1 & -1 & i & i & -1 \\
\hline
\end{array}

 

[fresh_42]

The real valued characters are $\chi_1,\chi_2,\chi_5,\chi_6$ where $\chi_1$ is the principal character. I think you left out a calculation step which is confusing here. It should be ...
\begin{align*}
A(225)&=\sum_{d|225}\chi_2(d)=2 [\chi_2(1)+\chi_2(3)+\chi_2(9)]+\chi_2(5)+\chi_2(13)+\chi_2(15)\\
&=2[1-1+\chi_2(9)]+\chi_2(5)+1-1\\
&=2\chi_2(9)+\chi_2(5)=3
\end{align*}
and the same for $\chi_5$ and $\chi_6.$