Problem: Show that is composite for all . What I have so far: In general, but this only holds for odd. Is there a way to factor when is even? Or would I have to find a solution specific to ?
Last edited by star_tenshi; January 22nd 2009 at 11:23 AM. Reason: switched odd / even for n. (>.<)
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Originally Posted by star_tenshi Problem: Show that is composite for all . What I have so far: In general, but this only holds for odd. Is there a way to factor when is even? Or would I have to find a solution specific to ? Note: You may want to wait for a more NT minded member to answer this in case of error Have you tried induction?
Originally Posted by star_tenshi Problem: Show that is composite for all . What I have so far: In general, but this only holds for odd. Is there a way to factor when is even? Or would I have to find a solution specific to ? You can use that identity with n=3, if you write 8 as 2^3.
Originally Posted by Opalg You can use that identity with n=3, if you write 8 as 2^3. Yes, I know that, but I am looking at the case where n is even. Even if I were to use , is not always odd.
Make use of the identity: Here imagine: and For all , we have: So what can we conclude?
OHHH! Now I see what Opalg was trying to convey. Thanks again o_O for clearing that up! We can conclude that is composite. Then to generalize, any number that can be expressed as is composite because it can be broken down into the factors !
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