Prove-that-n-4-4-n-is-composite-for-all-integer-values-of-n-greater-than-1-

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 .

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 = 2 m + 1 for some m ⩾ 1 .

Hence :

n^{4} + 4^{n} = {(n^{2})}^{2} + {(2^{n})}^{2} = {(n^{2} + 2^{n})}^{2} - 2 n^{2} 2^{n}

= {(n^{2} + 2^{n})}^{2} - 2^{n + 1} n^{2}

substituting n = 2 m + 1 :

n^{4} + 4^{n} = (n^{2} + 2^{2 m + 1}) - 2^{2 m + 2} n^{2} =

= {(n^{2} + 2^{2 m + 1})}^{2} - {(2^{m + 1} n)}^{2} =

= (n^{2} + 2^{2 m + 1} + 2^{m + 1} n) (n^{2} + 2^{2 m + 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^{2 m + 1} - 2^{m - 1} n =

= {(n - 2^{m})}^{2} + 2^{2 m} > 1 \Rightarrow

\Rightarrow n^{4} + 4^{n} is composite

Commented by Tinkutara last updated on 25/Sep/17

Thank you very much Sir!