Show that for any $n$, there are infinitely many cubes of the form $2^na - 9$.

Progress: We use induction on $n.$ For $n=1$ it works. Say it works for $n-1$. We will show for $n$. Note that $2^na-9$ is $$2^{n-1}2a-9.$$ If we have infinite $a$ such that $a$ is even for $n-1$, then we are done. Else, only finite amounts of $a$ are even.

So after a large constant $N$, we have $2^{n-1}a-9=x^3$ for only odd $a$.

So say $$2^{n-1}a_1-2^{n-1}a_2=u^3-w^3\implies 2^{n-1}|u^3-w^3$$

But I couldn't get anything further.

I thought about taking $v_2$, as in, if $2^na=x^3+9$ then we can take $x=y^2$. So we have $$2^na=y^2+3^2.$$


1 Answer 1


Suppose we have some positive $b$ for which $2^nb-9=u^3$ for some positive integer $u$. Notice that $u$ must be odd. Note then that \begin{equation} 2^n(b+3u^2+3u2^n+(2^{n})^2)-9=u^3+3u^22^n+3u(2^n)^2+(2^n)^3=(u+2^n)^3, \end{equation} and so $a=b+3u^2+3u2^n+(2^n)^2$ also satisfies the equation $2^na-9=v^3$ for some integer $v$.

However, $3u^2+3u2^n+(2^n)^2$ must be odd, as $u$ is odd. If $b$ was even, then $a$ is odd, and if $b$ was odd, then $a$ is even. Reiterating this argument, we find that if one solution $b$ exists, then we must in fact have an infinite amount of solutions, and that these solutions alternate signs, so that we also have an infinite amount of even solutions.

Your proof by induction then works.

  • $\begingroup$ Yes. Got it. Thankyou. $\endgroup$
    – Raheel
    Commented Oct 27, 2022 at 3:48

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