09-09-2009 Prove composite

Is it old ???
I dont know that..
just a try..I am not good at theory of numbers though..

I think there should be a condition taht n>1 becoz for n=1
it becomes 5 which is prime

anyways
if n is even
then 4ⁿ=even
and n⁴ is obviously even..so 4ⁿ+n⁴ is even
also it will be larger than 2....so it is a composite no.

Now If n is odd
4ⁿ+n⁴
if we put n=2m+1 in 4ⁿ
we get
4^2m4+n⁴
=>4.(2^m)⁴+n⁴
now putting 2^m=l
=>4l⁴+n⁴
=> (n²+2l²)²-(2nl)²
which is composite

Plz tell me of mistakes which I have committed...

@eureka 1 doubt
when n is odd we can surly express dat as 2m+1

so we get 4ⁿ+n⁴=4^2m+1+(2m+1)⁴

which is equal to
( 2^4m+2+2m(some constant)+1)......on binomially expandin (2m+1)⁴
now temme how did u proove this to be composite!!

When n is even and n>=2 then, the expression has a divisor 2.

When n is odd let n= 2k+1

n⁴+4ⁿ
=(2k+1)⁴+4^2k+1
=(2k+1)⁴+4.(2k)⁴
={(2k+1)²+2.2^2k}² - 4{(2k+1)2k}²
={(2k+1)²+2.2^2k+2.2^k(2k+1)}{(2k+1)²+2.2^2k-2.2^k(2k+1)}

Hence the expression is composite.
Am i right ? Any shortcut method ?

We know that for all nos N except of the form 4k+2 there exists integers x and y such that x^2-y^2=N
For some number representable as the difference of 2 squares - it can never represent a prime, as (x+y) or (x-y) or both shall divide it, meaning the above number cannot be a prime!

4ⁿ+[(n+1)(n-1)]²+1+2(n+1)(n-1)

which is odd clearly but dunno how to prove that the above no is composite.can u give me hints to prove from this step sir.

we can easily prove by binomial expansion.

How on earth can I miss this?????
(n⁴+4ⁿ)=(n²+2ⁿ)²-2ⁿ⁺¹n²
2ⁿ⁺¹n²=(2^a)²n²
I guess this finishes it!