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helpless06 · Dec 7, 2007, 06:41 AM

Show that: n^4+4 cannot be prime for n>3.

ebaines · Dec 7, 2007, 09:37 AM

You can rewrite the equation by completing the square, as follows:

$<br /> n^4+4\ =\ n^4 + (2\cdot n^2 \cdot 2 - 2\cdot n^2 \cdot 2) + 4 \ =\ (n^2 + 2)^2 - (2n)^2 \\<br />$

Now you have the difference of two squares, which can be written as the product of two terms. Can you take it from here? You'll find that for n>1, the two terms are each > 1, hence n^4+4 can not be prime for n>1.

cool_dude · Dec 7, 2007, 02:35 PM

Ah but the real proof lies in mathematical induction! At least that's the best way to proof this! Look at the previous post where you posted something similar on how to do mathematical induction. Try it out and show me the first few steps and I'll help you further.

ebaines · Dec 7, 2007, 03:34 PM

CD - help me out here -- I don't see how induction would work on this. Given that k^4+4 is a composite number (not prime), and then consider what happens if we replace k by (k+1), what you get is:

(k+1)^4 + 1 = (k^4+1) + (4k^3 + 6k^2 + 4k +1).

Now we know that k^4+1 is composite, but how does that help you prove that the sum of that composite plus the (4k^3 + 6k^2 + 4k +1) must be composite?

cool_dude · Dec 7, 2007, 04:05 PM

just out of curiousity how did u get 4k^3 + 6k^2 +4k +1 because I think that's wrong

Either way you know that 4k^3 is composite because 4 times an integer is composite, you know 6k^2 and 4k are composite for the same reasoning. Now I'm a little brain dead on the 1 lol. I'm not too sure on the 1 part at the moment maybe you can see better than me lol. If it wasn't for the 1 it would obviously be even.

ebaines · Dec 7, 2007, 04:32 PM

The problem is that the sum of composites may not necessarily be composite. For example, 4 + 9 = 13. This is why I don't see how induction would help with this problem.

As for my math, it's like this:

(k+1)^4 + 1 = (k^4 + 4K^3+6k^2+ 4k + 1) +1

= (k^4 + 1) + (4K^3+6k^2+ 4k + 1)

cool_dude · Dec 7, 2007, 06:16 PM

why are you doing +1? You say (k+1)^4+1? But should it not be (k+1)^4+4??

Also I just see where I misunderstood the problem thanks for pointing it out. See I thought u had to prove that it will be even but that is not the case since 9 isn't prime nor even. So offcourse my way doesn't work in this case. But if you had to strictly prove that the sum was even than my way should work.

ebaines · Dec 8, 2007, 08:02 AM

Originally Posted by cool_dude

why are you doing +1? you say (k+1)^4+1? but should it not be (k+1)^4+4???

You're right, sorry. I guess I must have been thinking about another problem...

So the math goes like this:
Given k^4+4 = composite, then
(k+1)^4+4 = (k^4 + 4k^3 + 6k^2 + 4k +1) + 4 = (k^4+4)+(4k^3 + 6k^2 + 4k +1)
= composite + 4k^3 + 6k^2 + 4k +1

I still don't see how this line of reasoning will get to an eventual proof.

cool_dude · Dec 8, 2007, 09:00 AM

See the problem with this question is that I didn't take into consideration all non prime numbers. Just when I saw this question it looked like mathematical induction proof because I thought oh if its not prime then its even. Forgot to take into consideration all composite numbers such as 9. so that kind of screws up my whole proof.