2nalfq8

2022-07-16

Is $g\left(x,y\right)=\frac{f\left({x}^{2y+1},y\right)}{f\left(x,y\right)}$ always an integer?

Aryanna Caldwell

Expert

If y is even, then
$g\left(x,y\right)=\frac{\frac{{x}^{y\left(2y+1\right)}-1}{{x}^{2y+1}+1}}{\frac{{x}^{y}-1}{x+1}}=\frac{\left({x}^{y\left(2y+1\right)}-1\right)\left(x+1\right)}{\left({x}^{2y+1}+1\right)\left({x}^{y}-1\right)}.$
If y is odd, then
$g\left(x,y\right)=\frac{\frac{{x}^{y\left(2y+1\right)}-1}{{x}^{2y+1}-1}}{\frac{{x}^{y}-1}{x-1}}=\frac{\left({x}^{y\left(2y+1\right)}-1\right)\left(x-1\right)}{\left({x}^{2y+1}-1\right)\left({x}^{y}-1\right)}.$
One has
${x}^{t}-1=\prod _{d\mid t}{\mathrm{\Phi }}_{d}\left(x\right),\phantom{\rule{2em}{0ex}}{x}^{t}+1=\prod _{\begin{array}{c}d\mid 2t\\ d\nmid t\end{array}}{\mathrm{\Phi }}_{d}\left(x\right).$
As a result, for example, using that $gcd\left(y,2y+1\right)=1$ you can show that for y odd

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