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

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 $$2^na=y^2+3^2.$$

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 $$$$2^n(b+3u^2+3u2^n+(2^{n})^2)-9=u^3+3u^22^n+3u(2^n)^2+(2^n)^3=(u+2^n)^3,$$$$ and so $$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.