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Sum of two powers (Posted on 2004-08-12) Difficulty: 4 of 5
If n is an integer, show that n4 + 4n is never a prime for n>1.

See The Solution Submitted by Federico Kereki    
Rating: 4.2500 (4 votes)

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Solution Complete solution Comment 7 of 7 |

If n is even, then 2 divides n4 + 4n.
Clearly (n4 + 4n)/2 > 2, for n > 1.
This establishes the result for even n.

If n is odd, we will write n4 + 4n as the difference of two squares of integers, and hence obtain a factorization.  Setting n = 2m + 1, we have

<table><tbody><tr class="h"><td class="u">n4 + 42m+1</td><td> = (n2)2 + (22m+1)2</td></tr> <tr><td> </td><td> = (n2 + 22m+1)2 − 22m+2n2</td></tr><tr><td> </td><td> = (n2 + 22m+1 + 2m+1n)(n2 + 22m+1 − 2m+1n)</td></tr></tbody></table>

We must now show that both factors are greater than one.

Clearly n2 + 22m+1 + 2m+1n > 1, for n > 1.

<table><tbody><tr class="h"><td class="u">Next consider f(n,m)</td><td> = n2 + 22m+1 − 2m+1n</td></tr><tr><td> </td><td> = (n − 2m)2 + 22m</td></tr></tbody></table>

For n > 1 (m > 0), 22m > 1; hence f(n,m) > 1 for all odd n > 1.

Therefore n4 + 4n is composite for all integers n > 1.


  Posted by Danish Ahmed Khan on 2012-10-24 14:42:01
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