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Prove number is composite?
Prove that n^4+4^n is composite if n is an integer >= 2
1 Answer
- \/::/\Lv 41 decade agoFavorite Answer
Haha! nice to see this problem! n^4+4^n=n^4+4(4^(n-1))=n^4+4(2^(2n-2)).
Now if n is even it clearly is composite since it is even. If not then n is odd, and 2n-2 is divisible by 4. Let 2n-2=4k.
Then n^4+4^n=n^4+4(2^(4k). Now this factorises as the Sophie Germain's identity, look here http://www.artofproblemsolving.com/Wiki/index.php/...
Both factors are greater than one, so n^4+4^n is composite.
