# (n^4) + 4 is composite ?

1. Apr 19, 2004

### Gokul43201

Staff Emeritus
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 ?

2. Apr 19, 2004

### Hurkyl

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

3. Apr 20, 2004

### cragwolf

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.

4. Apr 20, 2004

### Gokul43201

Staff Emeritus
thanks

thanks crag,
I must be blind !

5. Apr 20, 2004

### cragwolf

On no! The blind leading the blind!

6. May 27, 2004

### yrch

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

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

7. May 27, 2004

### arildno

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

8. May 27, 2004

### Gokul43201

Staff Emeritus
thanks, yrch...that's nice.

Now I really feel like \$#!+.

Last edited: May 27, 2004