Obviosly not, by Dirichlet theorem. If $a_n=a_0+nh=gcd(a_0,h)b_n$, $b_n$ is prime $p>gcd(a_0,h)$ for infinetely many $n$.

WakeUp wrote: Does there exist an infinite non-constant arithmetic progression, each term of which is of the form $a^b$, where $a$ and $b$ are positive integers with $b\ge 2$? It is enough to show that the numbers of the form $a^b$ become arbitrarily sparse. That is, letting $p(n)$ denote the number of such numbers less than or equal to $n$, we show that $\frac{p(n)}{n}$ can be made arbitrarily small. Indeed, let n = $2^k$. Then the exponents b that contribute terms $a^b$ counted in $p(n)$ satisfy $b \leq k$. The number of terms each exponent contributes is at most $2^{\frac{k}{2}}$, so we have $\frac {p(2k)}{2k} \leq \frac{k 2^{\frac{k}{2}}}{2^k} = \frac {k 2^k}{2}$ ,which clearly approaches 0 as $k \rightarrow \infty$.

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Correcting the typos ... so we have $\dfrac {p(2^k)}{2^k} \leq \frac{k 2^{\dfrac{k}{2}}}{2^k} = \frac {k} {2^{\dfrac{k}{2}}}$ ,which clearly approaches 0 as $k \rightarrow \infty$. Interestingly enough, there are finite, as long as wanted, arithmetic progressions made of perfect powers, see e.g. [W. Sierpinsky, 250 Number Theory Problems].

mavropnevma wrote: Correcting the typos ... so we have $\dfrac {p(2^k)}{2^k} \leq \frac{k 2^{\dfrac{k}{2}}}{2^k} = \frac {k} {2^{\dfrac{k}{2}}}$ ,which clearly approaches 0 as $k \rightarrow \infty$. Interestingly enough, there are finite, as long as wanted, arithmetic progressions made of perfect powers, see e.g. [W. Sierpinsky, 250 Number Theory Problems]. yes, $\frac{k 2^{\dfrac{k}{2}}}{2^k} = \frac {k} {2^{\dfrac{k}{2}}}$ ,not mine

to say more clearly,if $(a,d)=1$,then there exists a prime in it,withered. let$(a,d)=g$ then in$\frac{a}{g} n+\frac{d}{g}$there exists a prime sufficiently large ,it isnt' a divisor of $g$ again contradiction.

Yet another one, letting $a_n=a_0+dn$, where $d$ is the common difference, it suffices to establish the existence of a prime $p$, and an integer $n$ such that $a_n$ is divisible by $p$, but not $p^2$. Now, fix a prime $p>a_0,d$ and observe that $a_0+dn\equiv p\pmod{p^2} \iff n\equiv d^{-1}(p-a_0)\pmod{p^2}$, where $d^{-1}$ is the multiplicative inverse of $d$ in modulo $p^2$ (which exists, as $p>d$ implies $(p^2,d)=1$).