Let $p$ is give postive integers, show that:there exist infinitely many integers $q$ such that $3^{q-1}\cdot q\cdot p$ is cubic number?

Alternatively, I considered the form $q=p^2\cdot l^3$?But How to prove there exist infinitely $l$ such $p^2\cdot l^3-1\equiv0\pmod 3?$I can't figure it out

  • $\begingroup$ If $p$ is not divisible by 3, then $p^2 \equiv 1$ (mod 3), so any $l$ of the form $3k+1$ will satisfy your condition. However, if $p$ is divisible by 3, $p^2l^3-1$ will never be equal to 0 mod 3. You could try incorporate the factors of 3 in $p$ into the $3^{q-1}$. $\endgroup$
    – Maciek
    Sep 7 '15 at 8:25


Assume first that $3\not\mid p$.

Set $q=3k$. (If $3\mid p$ but $9\not \mid p$, set $q=3k+1$, while if $9\mid p$ but $27\not \mid p$ set $q=3k+2$, etc). Then $$ N=3^{q-1}qp=(3^{k})^3kq $$ It thus suffices to show that there exist infinitely many $k$, such that $pk$ is a cubic number.

  • $\begingroup$ Nice,But How to show that the following there exist infinitely many $k$ such $p(3k+1)$ is cubic number? $\endgroup$
    – user223800
    Sep 7 '15 at 8:32
  • $\begingroup$ "Hint" should not refer to an idea useful to solve a part of the problem but to indications sufficient to solve it entirely. Here you reduced problem P asked by the OP to some question Q. Is the answer to Q, obvious? Is Q even significantly simpler than P? $\endgroup$
    – Did
    Sep 7 '15 at 8:45
  • $\begingroup$ @Australia See my updated answer $\endgroup$ Sep 7 '15 at 8:56

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