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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 answer | Comment 4 of 7 |

As said below, we only need to consider the odd n's.  Let n=2k+1, where k is a positive integer.

Then

n4 + 4n = n4 + 42k+1
n4+4(2k)4

Now if we can factor this we are done.  This expression has the form a4+4b4, where a = n, b = 2k.  But
a4+4b4 = (a2+2ab+2b2)(a2-2ab+2b2)

Therefore,

n4+4(2k)4 = [(n2+2(n)(2k)+2(2k)2]x[(n2-2(n)(2k)+2(2k)2]

Now it remains to check that both factors are not 1.  Notice the first term is always bigger than 1 as n>1, k>=1.  The second term can be reexpressed as

(n2-2(n)(2k)+2(2k)2 = n2-n2k+1+22k+1
= n2+2k+1(2k-n)

If 2k-n = 2k-2k-1 is positive then we are done.  However, simply notice for k>=3, it is indeed positive.  So we'll need to check for k=1 and k=2.

But for k = 1 and k=2, or equivalently, n=3 and n=5,
n4 + 4n = 145 and 1649 respectively, which can be factored by 5 and 17 respectively.  Therefore we are done.

Edited on August 12, 2004, 3:35 pm
  Posted by Bon on 2004-08-12 15:31:37

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