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Question Number 21423 by Tinkutara last updated on 23/Sep/17

$$\mathrm{Prove}\mathrm{that}{n}^4+4^n\mathrm{is}\mathrm{composite}\mathrm{for}\mathrm{all}$$$$\mathrm{integer}\mathrm{values}\mathrm{of}n\mathrm{greater}\mathrm{than}1.$$

Answered by dioph last updated on 24/Sep/17

$$\mathrm{If}n\mathrm{is}\mathrm{even},\mathrm{both}{n}^4\mathrm{and}{4}^n\mathrm{are}\mathrm{even}$$$$\mathrm{so}2\mid n^4+4^n.$$$$\mathrm{If}n\mathrm{is}\mathrm{odd},n=2m+1\mathrm{for}\mathrm{some}m⩾1.$$$$\mathrm{Hence}:$$$$n^4+4^n={\left(n^2\right)}^2+{\left(2^n\right)}^2={(n^2+2^n)}^2-2n^2{2}^n$$$$={(n^2+2^n)}^2-2^{n+1}{n}^2$$$$\mathrm{substituting}n=2m+1:$$$$n^4+4^n=(n^2+2^{2m+1})-2^{2m+2}{n}^2=$$$$={(n^2+2^{2m+1})}^2-{\left(2^{m+1}n\right)}^2=$$$$=(n^2+2^{2m+1}+2^{m+1}n)(n^2+2^{2m+1}-2^{m+1}n)$$$$\mathrm{As}{n}^4+4^n\mathrm{is}\mathrm{positive}\mathrm{and}\mathrm{the}\mathrm{first}$$$$\mathrm{factor}\mathrm{above}\mathrm{is}\mathrm{positive},\mathrm{the}\mathrm{second}$$$$\mathrm{is}\mathrm{positive}\mathrm{as}\mathrm{well}.\mathrm{It}\mathrm{is}\mathrm{easy}\mathrm{now}$$$$\mathrm{to}\mathrm{see}\mathrm{that}{n}^2+2^{2m+1}-2^{m-1}n=$$$$={(n-2^m)}^2+2^{2m}>1\Rightarrow$$$$\Rightarrow n^4+4^n\mathrm{is}\mathrm{composite}$$

Commented by Tinkutara last updated on 25/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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