Show that only positive integer value of $a$ for which $$4(a^n+1)$$ is a perfect cube for all positive integers $n$, is $1$. Rewriting the equation we obtain: $$4(a^n+1)=k^3$$ It is obvious that $k$ is even and that in his prime factorization presents a prime factor $=2$ I proved to solve with LTE and other methods but in vain. How can I solve it?
Edit: Supposing that a prime factor of $k$ ($k=p_1^{3q}\cdot p_2^{3q}\cdot ....\cdot p_j^{3q}$) is a divisor of $a+1$ (i.e. $p|a+1$) therefore the greatest power of $p$ that divides $a^n+1$ has to be a multiple of $3$: $$\upsilon_p(a+1)+\upsilon(n)=3q$$ but we can note that $3q$ depends not only by $a$ but also by $n$ therefore this isn't possible for hypothesis. But if $p$ isn't a divisor of $a+1$, How can I continue?
No need of LTE. Suppose $$4(a^n+1)=k^3 \tag{$\star$}$$ for some $a>1$. Then, since $a^n+1>2$, we must have $$a^n+1=16b$$ $$ a^n=16b-1 \tag{P(n)}$$ for some $b\ge1$ (in fact $b$ is a cube, but it is an unnecessary information for our purposes). However, if $P(n)$ holds, then multiplying both sides by $a$ we see that $P(n+1)$ also holds if and only if there exists a positive integer $c$ such that $a-1=16c$:$$a^{n+1}=16ab-a \\ a^{n+1}=16ab-(a-1)-1 \\ a^{n+1}=16\left(ab-\frac{a-1}{16}\right)-1.$$ But this yields that $P(n)$ is equivalent to $$a^n-1=16b-2 \\ (a-1)\left(a^{n-1}+a^{n-2}+\cdots+1\right)=16b-2 \\ 16c\left(a^{n-1}+a^{n-2}+\cdots+1\right)=16b-2,$$ which is clearly impossible, hence we conclude $a=1$ alone satisfies $(\star)$ for all positive integers $n$.