Consider $n \ (>1)$ lotus leaves placed around a circle. A frog jumps from one leaf to another in the following manner. It starts from some selected leaf. From there, it skips one leaf in clockwise direction and jumps to the next one. Then it skips exactly $2$ leaves in clockwise direction and jumps to the next one. Then the frog skips $3$ in clockwise direction and so on. Suppose it turns out that if frog continues this way all leaves are visited by the frog. Show that $n$ cannot be odd.

My attempt,

Let frog be initially placed at leaf named $0$, and name the remaining leaves $1$, $2$ , $3$... $n - 1$ in clockwise direction.
After $k$ moves the frog will be on label number $0+(1+1)+(2+1)+\ ...\ + (k+1) = k (k+3)/2$.

Thus the question reduces to proving that there does not exist any $i$ such that $n = 2i+1$ and the congruence stated below is satisfied for all $t$ for some value of $k$:

$k(k+3)/2 \equiv t \,\pmod n, \;\; 0\leq t \lt n$