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Oct 28, 2017 at 14:47 comment added Wojowu @joro Yes, make it $g(x)=x^2+1,f(x)=x^2+1+(n^2+1)!$.
Oct 28, 2017 at 14:45 comment added joro Is it still easy if we require $g$ to be squarefree?
Oct 28, 2017 at 14:17 vote accept joro
Oct 28, 2017 at 14:13 history answered Wojowu CC BY-SA 3.0