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Jan 27, 2021 at 21:33 history edited T. Amdeberhan CC BY-SA 4.0
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Jan 26, 2021 at 23:23 comment added Will Sawin If $m$ is odd, then there is a trivial identity. $f_n(t^m)= (t^{m-1} - t^{m-1} + \dots + t^2 - t + 1)^n f_n(t)$. I don't see how you could have a meaningful non-trivial congruence while that identity exists. But maybe if $m$ is even you could do something cool.
Jan 26, 2021 at 23:04 comment added T. Amdeberhan @WillSawin: thanks, I fixed accordingly.
Jan 26, 2021 at 23:03 history edited T. Amdeberhan CC BY-SA 4.0
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Jan 26, 2021 at 22:22 comment added Will Sawin You want a congruence between $f_n ( t^m)$ and a multiple of $f_n(t)$?
Jan 26, 2021 at 22:13 history asked T. Amdeberhan CC BY-SA 4.0