An result : $ f(x)$ is a perfect square for all $ n\in N$ then $ f(x)$ is a square of a polynomial. Now consider our problem : $ a_{n_0+1}=p^3=an_0+b$ then Imply that $ (p+ta)^3$ is a term of this sequence. $ t\in N$ Imply that $ n_t=\frac{(p+ta)^3-b}{a}=f(t)$ Because $ f(t)$ is a polynomial of t with degree 3 . So it can not is a perfect square for all $ t\in N$ This problem claim.

If $ f(t_0)$ isn't a square then $ (p+t_0a)^3$ too. Why?

N.T.TUAN wrote: If $ f(t_0)$ isn't a square then $ (p + t_0a)^3$ too. Why? I don't understand ? I have prove that $ f(t)$ contain infinite term not is a perfect square.

Hint:

N.T.TUAN wrote: An infinite increasing arithmetical progression consists of positive integers and contains a perfect cube. Prove that this progression also contains a term which is a perfect cube but not a perfect square. arithmetical progression$ (ai + b)_{ i\in\mathbb{N}}$ we suppose that $ p^3 = an + b$ for some $ n\in\mathbb{N}$ then $ \forall t\in\mathbb{N}: \ (p + at)^3 = a(n + 3pat + 3p^2t + a^2t^3) + b\in\{ai + b: \ i\in\mathbb{N}\}$ if $ a^2a+p$ and $ a^2(1+a)+p$ are perfect squars then $ (a^3+a+p)-(a^3+p)\ge (a+1)^2-a^2$ (impossible) then $ \exists t\in\{a^2,a^2+1\}: \ p + at$ isn't perfect square. then $ (p + at)^3\in\{ai + b: \ i\in\mathbb{N}\}$ a perfect cube but not a perfect square.

I will give a complete solution to this problem: Let $ A$ be an arithmetical progression and $ a^3\in A$.Consider 2 cases: 1.$ a^3$ is not a perfect square,hence we are done. 2.$ a^3$ is a perfect square then using the lemma(see my post above),we obtain that $ a = b^2$,for some $ b\in N$. Also we know that $ (a + ad^2)^3\in A$,hence if $ (a + ad^2)$ is a perfect square too,then $ b^2 + b^2d^2$ is a perfect square too,which is impossible.Contradiction.

Sorry I don't read problem carefully . I have prove that there exist infinite $ n$ such that $ a_n$ is a perfect cube but n is not a square. This problem can be general as follow : Problem 1 Let $ p,q\in N$ where $ q>1$ and $ \gcd(p,q)=1$ Prove that if a arithmetic sequence $ a_n$ contain an $ p$ th power then it contain infinite term be a p th power but not a $ q^{th}$ power of integer . Proof :$ x_{n+1}=an+b$ Let $ k^p$ is a term of this sequence then $ (k+tb)^p$ also is a term too. Let chose $ k+tb$ is not $ q^{th}$ power then it is not a $ q^{th}$ power . This problem was be claim.