24
eureka123
·2009-09-14 10:44:23
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
42m4+n4
=>4.(2m)4+n4
now putting 2m=l
=>4l4+n4
=> (n2+2l2)2-(2nl)2
which is composite
Plz tell me of mistakes which I have committed...
1
archana anand
·2009-09-14 11:13:14
@eureka 1 doubt
when n is odd we can surly express dat as 2m+1
so we get 4n+n4=42m+1+(2m+1)4
which is equal to
( 24m+2+2m(some constant)+1)......on binomially expandin (2m+1)4
now temme how did u proove this to be composite!!
1
Bicchuram Aveek
·2009-09-14 12:15:57
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+42k+1
=(2k+1)4+4.(2k)4
={(2k+1)2+2.22k}2 - 4{(2k+1)2k}2
={(2k+1)2+2.22k+2.2k(2k+1)}{(2k+1)2+2.22k-2.2k(2k+1)}
Hence the expression is composite.
Am i right ? Any shortcut method ?
24
eureka123
·2009-09-14 12:28:41
@ archna..I did that way becoz only 3 days back I read about Sophie Germain's Identity [3]
so I tried to bring the question to that form.......I dont think I commited any mistake in it...but still....i need experts approval..
62
Lokesh Verma
·2009-09-15 01:02:29
do you need to take the cases even and odd???
This is much more simple? or is it?
62
Lokesh Verma
·2009-09-15 01:30:16
Hint: Can you express the above as the difference of 2 squares?
11
Devil
·2009-09-15 02:23:18
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!
62
Lokesh Verma
·2009-09-15 02:26:43
soumik but that does not prove that N is prime..
what if x-y is 1 and x+y is prime
like 42-32=7?
3
msp
·2009-09-15 05:35:39
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.
24
eureka123
·2009-09-15 07:03:23
but u agree naa that n >1
11
Devil
·2009-09-15 10:22:39
How on earth can I miss this?????
(n4+4n)=(n2+2n)2-2n+1n2
2n+1n2=(2a)2n2
I guess this finishes it!