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Question Number 21423 by Tinkutara last updated on 23/Sep/17
Prove that n^4 + 4^n is composite for all integer values of n greater than 1.
$$\mathrm{Prove}\:\mathrm{that}\:{n}^{\mathrm{4}} \:+\:\mathrm{4}^{{n}} \:\mathrm{is}\:\mathrm{composite}\:\mathrm{for}\:\mathrm{all} \\ $$$$\mathrm{integer}\:\mathrm{values}\:\mathrm{of}\:{n}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{1}. \\ $$
Answered by dioph last updated on 24/Sep/17
If n is even, both n^4 and 4^n are even so 2 ∣ n^4 +4^n . If n is odd, n = 2m+1 for some m ≥ 1. Hence: n^4 +4^n = (n^2 )^2 +(2^n )^2 = (n^2 +2^n )^2 −2n^2 2^n = (n^2 +2^n )^2 −2^(n+1) n^2 substituting n = 2m+1: n^4 +4^n = (n^2 +2^(2m+1) )−2^(2m+2) n^2 = = (n^2 +2^(2m+1) )^2 −(2^(m+1) n)^2 = = (n^2 +2^(2m+1) +2^(m+1) n)(n^2 +2^(2m+1) −2^(m+1) n) As n^4 +4^n is positive and the first factor above is positive, the second is positive as well. It is easy now to see that n^2 +2^(2m+1) −2^(m−1) n = = (n−2^m )^2 +2^(2m) > 1 ⇒ ⇒ n^4 +4^n is composite
$$\mathrm{If}\:{n}\:\mathrm{is}\:\mathrm{even},\:\mathrm{both}\:{n}^{\mathrm{4}} \:\mathrm{and}\:\mathrm{4}^{{n}} \:\mathrm{are}\:\mathrm{even} \\ $$$$\mathrm{so}\:\mathrm{2}\:\mid\:{n}^{\mathrm{4}} +\mathrm{4}^{{n}} . \\ $$$$\mathrm{If}\:{n}\:\mathrm{is}\:\mathrm{odd},\:{n}\:=\:\mathrm{2}{m}+\mathrm{1}\:\mathrm{for}\:\mathrm{some}\:{m}\:\geqslant\:\mathrm{1}. \\ $$$$\mathrm{Hence}: \\ $$$${n}^{\mathrm{4}} +\mathrm{4}^{{n}} \:=\:\left({n}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{2}^{{n}} \right)^{\mathrm{2}} \:=\:\left({n}^{\mathrm{2}} +\mathrm{2}^{{n}} \right)^{\mathrm{2}} −\mathrm{2}{n}^{\mathrm{2}} \mathrm{2}^{{n}} \\ $$$$=\:\left({n}^{\mathrm{2}} +\mathrm{2}^{{n}} \right)^{\mathrm{2}} −\mathrm{2}^{{n}+\mathrm{1}} {n}^{\mathrm{2}} \\ $$$$\mathrm{substituting}\:{n}\:=\:\mathrm{2}{m}+\mathrm{1}: \\ $$$${n}^{\mathrm{4}} +\mathrm{4}^{{n}} \:=\:\left({n}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}{m}+\mathrm{1}} \right)−\mathrm{2}^{\mathrm{2}{m}+\mathrm{2}} {n}^{\mathrm{2}} \:=\: \\ $$$$=\:\left({n}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}{m}+\mathrm{1}} \right)^{\mathrm{2}} −\left(\mathrm{2}^{{m}+\mathrm{1}} {n}\right)^{\mathrm{2}} \:= \\ $$$$=\:\left({n}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}{m}+\mathrm{1}} +\mathrm{2}^{{m}+\mathrm{1}} {n}\right)\left({n}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}{m}+\mathrm{1}} −\mathrm{2}^{{m}+\mathrm{1}} {n}\right) \\ $$$$\mathrm{As}\:{n}^{\mathrm{4}} +\mathrm{4}^{{n}} \:\mathrm{is}\:\mathrm{positive}\:\mathrm{and}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{factor}\:\mathrm{above}\:\mathrm{is}\:\mathrm{positive},\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{is}\:\mathrm{positive}\:\mathrm{as}\:\mathrm{well}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{now} \\ $$$$\mathrm{to}\:\mathrm{see}\:\mathrm{that}\:{n}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}{m}+\mathrm{1}} −\mathrm{2}^{{m}−\mathrm{1}} {n}\:= \\ $$$$=\:\left({n}−\mathrm{2}^{{m}} \right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}{m}} \:>\:\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow\:{n}^{\mathrm{4}} +\mathrm{4}^{{n}} \:\mathrm{is}\:\mathrm{composite} \\ $$
Commented by Tinkutara last updated on 25/Sep/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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