Peter wrote: Show that if an infinite arithmetic progression of positive integers contains a square and a cube, it must contain a sixth power. http://www.kalva.demon.co.uk/short/soln/sh9715.html

Let $ r,s$ be coprime. If an arithmetic progression contains a $ r$-th power and a $ s$-th power, then it contains a $ rs$-th power. Proof: This is equivalent to: Let $ a,d$ be integers. If there are integers $ x,y$ such that $ a \equiv x^r \equiv y^s \mod d$, then there is an integer $ z$ such that $ a \equiv z^{rs} \mod d$. First observation: we can assume that $ d=p^q$ is a prime power, $ p$ prime. Indeed, by the chinese remainder theorem, an integer polynomial equation can be solved $ \mod n$ iff it can be solved $ \mod$ all prime powers dividing $ n$. Here, the equations are $ x^r-a , y^s-a, z^{rs}-a$. Second observation: let $ d=p^q$ as said, then we can assume $ \gcd(a,p^q)=1$. Indeed, if $ p|a$, we write $ x= p^u x', y= p^v y'$ with $ x', y'$ both not divisible by $ p$. Then $ x^r \equiv y^s \mod p^q$ gives $ a \equiv p^{ur} x'^r \equiv p^{vs} y'^s \mod p^q$. Thus either $ a \equiv 0 \mod p^q$, in this case we can take $ z=0$, or we get that $ vs=v_p(x^r) = v_p(a) = v_p(y^s) = ur <q$. This gives that $ r|v$, thus $ vs=rsw$ for some $ w$, giving that we simply want to solve $ z'^{rs} \equiv \frac{a}{p^{rsw}} \mod p^{q-rsw}$ with $ \frac{a}{p^{rsw}}$ coprime to $ p$. Third observation: the problem's statement is true if $ a$ is coprime to $ d$. Indeed, if $ x^r \equiv a \equiv y^s \mod p^s$ and $ p \nmid x,y$, we write $ 1=ur+vs$ for some integers $ r,s$. Then we take $ z=x^v y^u$ because then $ z^{rs} = x^{vrs} y^{urs} \equiv a^{vs+ur} = a \mod d$. Putting all together, we solved the problem.