Let $p$ be a prime number.($p \neq 2,3,5$)
Let $t^+,t^-,a$ be sequences.
$t^+_{k+5}=t^+_k,t^+_1=0,t^+_2=-1,t^+_3=-1,t^+_4=0,t^+_5=2$
$t^-_{k+5}=t^-_k,t^-_1=-1,t^-_2=0,t^-_3=0,t^-_4=-1,t^-_5=2$
$a_k=2(\cos\frac{k}{3}\pi-\cos\frac{k}{2}\pi)$
Then,
case $p\equiv1,4\pmod5$
$\underset{1\leq k\leq{p-1}}{\sum}\frac{a_k t^+_{p-k}}{k}\equiv0\pmod p$
case $p\equiv2,3\pmod5$
$\underset{1\leq k\leq{p-1}}{\sum}\frac{a_k t^-_{p-k}}{k}\equiv0\pmod p$
I have checked this for $p<10000$.
Can anyone prove this?
Note
$t^+_k-t^-_k=(\frac{k}{5})$
$t^+_k+t^-_k=e^{\frac{2}{5}k\pi i}+e^{\frac{4}{5}k\pi i}+e^{\frac{6}{5}k\pi i}+e^{\frac{8}{5}k\pi i}$