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

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(4^n-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.