An infinite arithmetic progression, whose terms are all positive integers, contains
(i) a perfect square which is not a perfect cube, and
(ii) a perfect cube which is not a perfect square.
Prove that the arithmetic progression contains a perfect sixth power.
Finding examples where the base is the same is trivial but to find a general proof I tried looking at examples where square and cube are co-prime (hence common difference d is also co-prime with these) and using mod d and Euler's theorem to come up with a relatively short algebraic proof however, nothing has immediately opened up.
We know that there exist $x$, $y$ with $x^2\equiv y^3\equiv a\pmod n$. Let $d$ be the order of $a$ mod $n$.
If $d=1$, we are done as $x^6\equiv1\pmod n$. If $n=2$, then $(xy)^6\equiv a^5\equiv a\pmod n$, and again we are done.
So assume $d>2$. For any $i,j$ we have $(x^iy^j)^6\equiv a^{3i+2j}\pmod n$. By the Chicken McNugget theorem, there exist $i,j$ such that $3i+2j=m$ for all $m>6-2-3=1$. Since $d-1>1$, we can find $i,j$ such that $3i+2j=d-1$, so $$(x^iy^j)^6\equiv a^{d-1}\pmod n\implies\left(x^{i(d-1)}y^{j(d-1)}\right)^6\equiv a^{(d-1)^2}\equiv a\pmod n,$$ and we are done!