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# Prove that for any integer n > 1,4^n + n^4 is composite.

Swapnil Saxena
102 Points
9 years ago

The question can be solved with sophie germain identity.

To prove that a no is non prime, we have to prove tht it can be resolutted into two integral factors

For odd n:

4n+n4=  4(4n-1)+n4 = 4(4(n-1)/4)4+n4. Taking 4(n-1)/4 as t , the equation becomes 4t4+ n4

Solving this by Sophie Germain identity

= (n4+2(t)2+2nt)(n4+2(t)2-2nt)

Now Putting 4(n-1)/4 in the palce of n , the equation becomes

=(n4+2(4(n-1)/2) +2n(4(n-1)/4))(n4+2(4(n-1)/2) -2n(4(n-1)/4 ))

=(n4+2n +2n(2(n-1)/2))(n4+2n -2n(2(n-1)/2 ))

Now fr proving that it is not a prime,  We have to prove that the two factors n4+2n -2n(2(n-1)/2 ) and n4+2n +2n(2(n-1)/2 )is integral. Since n4+2n is always integral we ave to prove 2n(2(n-1)/2)

For n is odd the no greater than 1 , the factors are definitely integers as (2(n-1)/2 ) is integral fr every odd n.

For even n:

Let n is a even no, then it can be represented in form of 2m where m is a natural no

(2m)4+(4)2m always consist of 2 as its factor. So it is a composite for even n.

Hence Proved.