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Question Number 65214 by naka3546 last updated on 26/Jul/19

$$Proveordisprovethat{2}^{101}\mid n^n-101.$$

Commented by naka3546 last updated on 26/Jul/19

$$npositiveinteger$$

Commented by Rasheed.Sindhi last updated on 27/Jul/19

$$Forn=1,n^n-101=1^1-101=-100$$$$And{2}^{101}\nmid -100.$$

Commented by naka3546 last updated on 27/Jul/19

$$anynthatsatisfyon?$$

Answered by MJS last updated on 27/Jul/19

$$n^n-101⩾2^{101}\Rightarrow n⩾2^{101}+101=2535301200456458802993406410853$$$$$$$$n^n-101=m\times 2^{101}$$$$n^n-101=2m\times 2^{100}$$$$n^n-101=2l$$$$n^n=2l+101$$$$n^n=2(l+50)+1$$$$n^n=2k+1\Rightarrow {\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{true}\mathrm{for}\mathrm{even}n$$$$$$$$2^{101}-101=7^2\times 263201\times 196582994842884397808597=$$$$=p^{2}qr$$$$n^n=m\times p^{2}qr$$$$m=\frac{n^n}{p^{2}qr}=\frac{{\left(p^{\alpha }{q}^{\beta }{r}^{\gamma }\right)}^{\left(p^{\alpha }{q}^{\beta }{r}^{\gamma }\right)}}{p^{2}qr}=$$$$=p^{\alpha p^{\alpha }{q}^{\beta }{r}^{\gamma }-2}{q}^{\beta p^{\alpha }{q}^{\beta }{r}^{\gamma }-1}{r}^{\gamma p^{\alpha }{q}^{\beta }{r}^{\gamma }-1}$$$$\mathrm{the}\mathrm{smallest}\mathrm{number}m\mathrm{is}\mathrm{the}\mathrm{one}\mathrm{with}$$$$\alpha =\beta =\gamma =1$$$$m_{min}=p^{pqr-2}{q}^{pqr-1}{r}^{pqr-1}\approx {10}^{\left({10}^{31}\right)}$$

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