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Question Number 193886 by Rupesh123 last updated on 22/Jun/23

Answered by MM42 last updated on 22/Jun/23

$$a=521({521}^n-{521}^{n-1}+1)=521m$$$$‘‘{521}^{″}isprimenumber.therefore‘‘m^{″}$$$$mustbeamultiple‘‘{521}^{″}.$$$$whichisvalidonlyfor‘‘n=1^{″}$$

Commented by Rupesh123 last updated on 22/Jun/23

Perfect ��

Answered by Subhi last updated on 22/Jun/23

$$521((521-1){521}^{n-1}+1)$$$$521({521}^n-{521}^{n-1}+1)$$$$tobeaperfectsquare$$$${521}^n-{521}^{n-1}+1={521}^{2m+1},wherem⩾0$$$$n⩾1$$$$or{521}^n-{521}^{n-1}+1={521}^{2k-1},wherek⩽0$$$$n⩾1$$$${521}^{2m+1}-{521}^n+{521}^{n-1}=1$$$$521({521}^{2m}-{521}^{n-1}+{521}^{n-2})=1$$$${521}^{2m}-{521}^{n-1}+{521}^{n-2}=\frac{1}{521}={\left(521\right)}^{-1}$$$$\therefore n-2⩽0\Rrightarrow n⩽2$$$$n-1⩽0\Rrightarrow n⩽1$$$$m⩽0$$$$\therefore m=0,n=1$$$${521}^{2k-1}+{521}^{n-1}-{521}^n=1$$$$521({521}^{2k-2}+{521}^{n-2}-{521}^{n-1})=1$$$${521}^{2k-2}+{521}^{n-2}-{521}^{n-1}=\frac{1}{521}$$$$n-1⩽0\Rrightarrow n⩽1\Rrightarrow n=1\Rrightarrow k=1$$

Commented by Rupesh123 last updated on 23/Jun/23

Perfect ��

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