What are the integers $a$ between 1 and 105 such that $35 \mid a^{3}-1$?
The cubes modulo 5 are, in order $0,1,3,2,4$ so $5 \mid a^{3}-1$ if and only if $a \equiv 1[5]$.

The cubes modulo 7 are, in order $0,1,1,-1,1,-1,-1$ so $7 \mid a^{3}-1$ if and only if $a \equiv 1,2,4[7]$. By the Chinese remainder theorem, there are therefore 3 possible remainders modulo 35 for $a$ which are $1,11,16$.

Consequently, the solutions are: $1,11,16,36,46,51,71,81$ and 86.