A curious sum for integers $\equiv 7\pmod 8$. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T10:19:14Z http://mathoverflow.net/feeds/question/120521 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120521/a-curious-sum-for-integers-equiv-7-pmod-8 A curious sum for integers $\equiv 7\pmod 8$. Roland Bacher 2013-02-01T14:13:51Z 2013-02-01T19:43:19Z <p>For $n$ a natural integer congruent to $7$ modulo $8$, one has seemingly always $$\sum_{k=1}^{(n-1)/2}\left(\frac{k}{n}\right)k=0$$ where $\left(\frac{k}{n}\right)$ denotes the Jacobi symbol.</p> <p>First cases:</p> <p>$n=7$: $1+2-3$</p> <p>$n=15$: $1+2+4-7$</p> <p>$n=23$: $1+2+3+4-5+6-7+8+9-10-11$</p> <p>I do not see any reason for this. Did I miss something obvious?</p> http://mathoverflow.net/questions/120521/a-curious-sum-for-integers-equiv-7-pmod-8/120524#120524 Answer by Abhinav Kumar for A curious sum for integers $\equiv 7\pmod 8$. Abhinav Kumar 2013-02-01T14:44:44Z 2013-02-01T19:43:19Z <p>There are probably lots of ways to see this, but here's one: let $S_1$ be the sum above, and let $$S_2 = \sum_{k=(n+1)/2}^{n-1} k \left( \frac{k}{n} \right).$$ Then $$S_1 + S_2 = S = \sum_{k=1}^{n-1} k \left( \frac{k}{n} \right).$$ Now you can rewrite $S$ (since $x \to 2x$ is a bijection mod $n$) as $$S = \sum_{k=1}^{(n-1)/2} 2k \left( \frac{2k}{n} \right) + \sum_{k=(n+1)/2}^{n-1} (2k-n) \left( \frac{2k}{n} \right) = 2S - n \sum_{k=(n+1)/2}^{n-1} \left( \frac{k}{n} \right),$$ where we used $(2/n) = 1$. Finally, we have $$S = 2S + n \sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right),$$ by changing $k$ to $n-k$ in the summation and using $(-1/n) = -1$. So $S = -n\sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right)$.</p> <p>On the other hand, we can switch the index in $S_2$ from $k$ to $n-k$ as well, to get $$S_2 = \sum_{k=1}^{(n-1)/2} (n-k) \left( \frac{n-k}{n}\right) = -n \sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right) + \sum_{k=1}^{(n-1)/2} k \left( \frac{k}{n} \right) = S + S_1 = 2S_1 + S_2.$$ (We used $(-1/n) = -1$ again).</p> <p>Therefore $S_1 = 0.$</p>