Proving (n^4) + 4 is Composite for All n>1: The Odd n Case

[Gokul43201]

How do you prove that (n^4) +4 is composite for all n>1 ?

I found this problem in a book I was reading. The even n part is obvious. The proof for odd n cannot be done by trying to factorize [(2k+1)^4]+4. I know this is true, because I've tried and moreover, the factors of the first few such numbers are : (2*2*5),(5*17),(2*2*5*13),(17*37),(2*2*5*5*13),(5*13*37),... The expected factors of 4 appear in the even numbers, but the factors for the odds seem patternless. So, I imagine that the proof must be done by some other means.

Any ideas ?

 

[Hurkyl]

Are there any factors that tend to occur frequently? Maybe you can solve the problem by cases.

 

[cragwolf]

(2*2*5),(5*17),(2*2*5*13),(17*37),(2*2*5*5*13),(5*13*37),... The expected factors of 4 appear in the even numbers, but the factors for the odds seem patternless.

Are you sure they're patternless? Just look at the odd n for now:

n=3: n^4+4 = 5*17
n=5: n^4+4 = 17*37
n=7: n^4+4 = 5*13*37 = 37*65

And I'll add n=9 for you:

n=9: n^4+4 = 5*13*101 = 65*101

See any pattern yet? That should lead you to the general proof.

 

[Gokul43201]

thanks

thanks crag,
I must be blind !

 

[cragwolf]

On no! The blind leading the blind!

 

[yrch]

how about:
$$n^4 + 4 = n^4 + 4 - 4n^2 + 4n^2 = (n^2 + 2)^2 - (2n)^2 = (n^2 + 2n + 2)(n^2 - 2n + 2)$$
:)

edit:
sorry, i pulled up an old post... just realized that it's from April not May 20th...

 

[arildno]

edit:
sorry, i pulled up an old post... just realized that it's from April not May 20th...

What a terrible misdemeanor on your part...
I am almost tempted not to welcome you

 

[Gokul43201]

thanks, yrch...that's nice.

Now I really feel like $#!+.