Question

how do i prove
n^4 + 4^n is never prime for n>1

Answer 1

Proof by induction? Show the statement holds for \(n=2\), then (assuming it's true for \(n=k\)) show that it holds for \(n=k+1\).

Answer 2

well yes I tried and am stuck there, where we prove for n=k+1

Answer 3

i have even tried this way,
clearly its true when n=2m for some natural no. m
but again stuck where we need to prove for n=2m+1

Answer 4

For the \(n=k+1\) case, you have \((k+1)^4+4^{k+1}\). Try expanding the binomial and see what happens for even and odd \(k\).

Answer 5

well i have assumed 4^k + k^4 = ab where a,b>1

now for k+1 part,

4^(k+1) + (k+1)^4
=> 3.4^k + ab + 4k^3 + 6k^2 + 4k +1

stuck here :/

Answer 6

Do you know about modular arithmetic?

Answer 7

bit

Answer 8

Clearly if n is even, the result will be divisible by 2. So you should only worry about showing that its not prime if n is odd. Modular arithmetic can take care of that case.

Answer 9

please guide me through ?
not completely, but hint me please

Answer 10

Try plugging in some odd numbers. Try to guess what always divides to the statement if n is odd.

Answer 11

well an interesting result that comes is that this f(n) is always multiple of 2 or more different primes

Answer 12

First note that every integer \(n>1\) is either even or odd.

If \(n\) is even, then \(n=2k\) for some positive integer \(k\), and thus \(n^4+4^n=2^4k^4+4^{2k}=2(2^3k^4+2^{4k-1})\), so clearly \(2\mid n^4+4^n\) and therefore \(n^4+4^n\) is not prime.

If \(n\) is odd, then \(n=2k+1\) for some positive integer \(k\). Consider \(4^n\mod5\) and it should be trivial.

Answer 13

what do we do about the polynomial ?

Answer 14

and yes i see by hit and trial that f(n) is always divisible by 5 !! :O

Answer 15

Euler's theorem tells us that \(4^n=4^{n\mod4}\mod5\). Consider, then:$$4^0=1\mod5\\4^1=4\mod5\\4^2=1\mod5\\4^3=4\mod 5$$Since \(n\) is odd, \(n=1,3\mod4\) and thus \(4^n=4\mod5\). This leaves us to show that \(n^4=1\mod5\).

Recognize $$n^4=(2k+1)^4=(2k)^4+4(2k)^3+6(2k)^8+4(2k)+1=1\mod5$$... and we're done.

Answer 16

not too sure how you directly conclude (2k)^4 +4(2k)^3 + 6(2k)^2 + 4(2k) is divisible by 5 ?

Answer 17

Can you show $$n^4-1=(n^2+1)(n^2-1)=(n^2+1)(n-1)(n+1)=0\mod5$$?

Answer 18

(2k)^2 is what I meant, btw.

Answer 19

try this also: for even numbers its true clearly...for odd numbers\[n^4+4^n=n^4+4^{2k+1}=n^4+4\times 4^{2k}=n^4+4\times (2^k)^4=n^4+4m^4\]try to factor last expression

Answer 20

oh of course!
(n^2 -2mn + 2m^2 ) (n^2 +2mn + 2m^2)

hence not prime !!

thank you @mukushla
and am sorry i am still unable to follow that mod5 method @oldrin.bataku :(

Answer 21

$$n+1=0\mod5\implies n=4\mod5\\n-1=0\mod5\implies n=1\mod 5\\n^2+1=0\mod5\implies n^2=4\mod5\implies n=2,3\mod 5$$So we've established $$n=1,3\implies n^4-1=0\mod5$$and thus $$n^4=1\mod 5$$

Answer 22

confused at n^2 + 1 = 0 part ?

Answer 23

oh no
got it

Answer 24

We factored \(n^4-1=(n^2+1)(n-1)(n+1)\) and showed that no matter what \(n\) is we end up with \(n^4-1=0\mod5\).

Answer 25

except for n=5m ?