$4(a^3+1)$ and $4(a^9+1)$ are both perfect cubes, hence their quotient $a^6-a^3+1$ is also a perfect cube. For $a>1$, $a^6-a^3+1<(a^2)^3$, hence $a^6-a^3+1\le(a^2-1)^3$, $3a^4-a^3-3a^2+2\le0$, $(a-1)(3a^3-2a^2-1)+1\le0$, a contradiction. Thus $a=1$.

Does anyone have a solution with LTE?

Suppose $a\neq 1$. If $a+1$ is not a power of $2$, let $p$ be an odd prime dividing $a+1$. Then by LTE $v_p(a^n+1)=v_p(a+1)+v_p(n)$ is a multiple of 3 for all odd positive integer $n$, clearly impossible. If $a+1$ is a power of $2$, Note that $v_2(4(a^3+1))=2+v_2(a+1)+v_2(3)=2+v_2(a+1)+v_2(1)=v_2(4(a+1))$, but $a^3+1>a+1$ implies there is an odd prime $p$ dividing $a^3+1$. Then by LTE $v_p(4(a^{3k}+1))=v_p(a^3+1)+v_p(k)$ is a multiple of $3$ for all odd integers $k$, thus contradiction and $a=1$.

I answer based on buzzychaoz's solution. We suppose that $a>1$.Obviously $2|a+1$.By Zsigmondy's theorem, ∃$p$(prime) s.t. $p|a^3+1$ and $p\nmid a+1\rightarrow p\neq 2$.By LTE $\forall k$(odd), $v_p(4(a^{3k}+1))=v_p(a^{3k}+1)=v_p(a^3+1)+v_p(k)$. $\forall k(odd):v_p(a^3+1)+v_p(k)\equiv 0\pmod{3}$ This is apparently absurd.$\blacksquare$

I have a same solution as jgnr also what is Zsigmondy's theorem

@fighter Brother! You should study Zsigmondy's theorem.This theorem kills many olympiad problem(number theory) instantly. See here http://www.math.ust.hk/excalibur/v16_n4.pdf

thank you

A Problem by DR. M. Jamaali

@math-helli Thanks,I see.

Takeya.O wrote: I answer based on buzzychaoz's solution. We suppose that $a>1$.Obviously $2|a+1$.By Zsigmondy's theorem, ∃$p$(prime) s.t. $p|a^3+1$ and $p\nmid a+1\rightarrow p\neq 2$.By LTE $\forall k$(odd), $v_p(4(a^{3k}+1))=v_p(a^{3k}+1)=v_p(a^3+1)+v_p(k)$. $\forall k(odd):v_p(a^3+1)+v_p(k)\equiv 0\pmod{3}$ This is apparently absurd.$\blacksquare$ why n=3k?

buiminhducduc wrote: Takeya.O wrote: I answer based on buzzychaoz's solution. We suppose that $a>1$.Obviously $2|a+1$.By Zsigmondy's theorem, ∃$p$(prime) s.t. $p|a^3+1$ and $p\nmid a+1\rightarrow p\neq 2$.By LTE $\forall k$(odd), $v_p(4(a^{3k}+1))=v_p(a^{3k}+1)=v_p(a^3+1)+v_p(k)$. $\forall k(odd):v_p(a^3+1)+v_p(k)\equiv 0\pmod{3}$ This is apparently absurd.$\blacksquare$ why n=3k? Now, natural number $n$ is arbitrary

Previous solutions are simpler but i will try to prove it with only one statement. $n=18$ so its going to be little harder. I will prove that the only natural solution for $n=18$ is $a=1$. On diophantine equation $4(a^{18}+1)=x^3$ $4(a^{18}+1)=x^3$ so set $x=2y$ $a^{18}+1=2y^3$ so $(a^6+1)(a^{12}-a^6+1)=2y^3$ obviously $a^{18}-a^6+1$ is odd so $a^6+1$ even. Also $(a^6+1,a^{12}-a^6+1)=3$ if $a^6=2mod3$ . If $a^6=0,1mod3$ then $(a^6+1,a^{12}-a^6+1)=1$ so we have 2 cases: First case $(a^6+1,a^{12}-a^6+1)=1$ so $a^6+1=2k^3$ $a^{12}-a^6+1=l^3$ $(k,l)=1$ , $(k,l)=y$ From second equation obiously $a$ bounded and finally we get $a=1$ Second case $(a^6+1,a^{12}-a^6+1)=3$ at this case notice that $a^{12}-a^6+1$ is never divisible by $9$ but divisible by $3$ so we get $a^6+1=2*9k^3$ $a^{12}-a^6+1=3l^3$ $(k,l)=1$ , $kl=y$ Now we see that the equation $a^6+1=2*9k^3$ has no solutions $mod3$ . So the only natural solutions of diophantine $4(a^{18}+1)=x^3$ are $(a,x)=(1,2)$ P.s Can you see why i tried to solve this problem for $n=18$ ?

if we take $n=3k$, we will have: $(-a^{k})^{3}- 2(-y)^{3}=1$ for some integer $y$, and by Skolem's theorem, $a$ must be $1$

Here is a quite different solution than all of the above. Clearly, the thesis of the problem implies that there exists a sequence $(k_n)_{n=1}^\infty$ of positive integers, such that $a^n+1 = 2k_n^3$ holds. Now, $a^{n+3}+1 = 2k_{n+3}^3$ implies, $\frac{a^3-1}{2}=(ak_n)^3 - k_{n+3}^3$. Now, if $a>1$, then the sequence $k_n$ is strictly increasing (to infinity). Moreover, $a^2k_n^2 + ak_n k_{n+3}+k_{n+3}^2\mid \frac{a^3-1}{2}$ also holds. Thus, if $a>1$, this divisibility implies a contradiction: $\frac{a^3-1}{2}$ is a fixed positive number, whereas $a^2k_n^2 + ak_n k_{n+3}+k_{n+3}^2\to+\infty$.

I hope this works, because this LTE solution seems shorter than the others posted.... If $a^2 + 1$ has an odd prime factor, say $p$, then LTE yields \[ \nu_p(a^{2k} + 1) = \nu_p(a^2 + 1) + \nu_p(k) \]whenever $k$ is odd; this is clearly not always divisible by $3$. Thus $a^2 + 1$ is a power of $2$. But it's also congruent to $2\pmod 4$, so $a = 1$.

Claim $(1)$: $a^3 + 1$ is a power of two. Proof: SFTSOC that some odd $p \in \mathbb{P}$ divides $a^3 + 1$. Then we also have $p \nmid a^3, 1$ and so (by LTE) for any odd $n \in \mathbb{Z}^+$,$$\nu_p\left(4\left(a^{3n} + 1^{3n}\right)\right) = \nu_p(4) + \nu_p\left(\left(a^3\right)^n + 1^n\right) = \nu_p\left(a^3 + 1\right) + \nu_p(n)$$Clearly there's some odd $n \in \mathbb{Z}^+$ for which the RHS is not divisible by $3$, but then $4(a^n + 1)$ isn't a perfect cube, a contradiction. So, $a^3 + 1 > 0$ is not divisible by any odd prime, which implies it is a power of two. Claim:\[a < 3 \qquad (2)\]Proof: SFTSOC that $a \geq 3$. Since $a^3 + 1 = (a + 1)(a^2 - a + 1)$, by $(1)$ both $a + 1$ and $a^2 - a + 1$ are powers of two. Together with the fact that $a^2 - a + 1 > a + 1$ (because $a(a - 2) > 0$), this means we can write$$a^2 - a + 1 = 2^i(a + 1)$$for some $i \in \mathbb{Z}^+$. Taking the above equation modulo $a$ gives $2^i \equiv 1 \pmod{a}$, and in particular $2^i > a$ (as $a > 1$ and $i > 0$). But then$$2^i(a + 1) > a^2 + a > a^2 - a + 1$$, a contradiction. Claim:\[a = 1 \qquad (3)\]Proof: If $a = 2$, then $4(a^1 + 1) = 12$ is not a perfect cube, a contradiction. So, $a \neq 2$, which means by $(2)$ that $a = 1$, as desired.