how would you prove that for 135 consecutive integers $a_1,a_2,...,a_{135}$ such that for all integers $k$, $1 \leq k \geq 135$ there exists a perfect cube $z_k > 1$ such that $z_k$ divides $a_k$?
Im stuck on this question and don't even know how to start. All I know is that if $z_k$ divides $a_k$ then $a_k = mod (z_k)$
If the question is what I have stated in my comment then, consider $135$ distinct primes $p_1,p_2, \ldots, p_{135}$. Now consider the following system of congruences: \begin{align*} x & \equiv 0 \pmod{p_1^3}\\ x & \equiv -1 \pmod{p_2^3}\\ x & \equiv -2 \pmod{p_3^3}\\ \vdots & \equiv \vdots\\ x & \equiv -134 \pmod{p_{135}^3} \end{align*} Then by the Chinese remainder theorem, there exists a solution $x_0$ to this system. Thus $\{x_0,x_0+1,x_0+2, \ldots, x_0+134\}$ will satisfy the given requirements.