iscomposite( n^n + 2*n + 1) always ?
in [Math]
|
Prev: X,Y Top. Spaces, f cont:f: X-- --Y is nullhomotopic if X is contractible
Next: Most *elementary* Q on quotient map of topological spaces
From: Christian Aebi on 5 Apr 2010 02:38 Many thanks to all those who supplied an answer (even 12 years ago!). No Kent, it wasn't homework and I should have thought a little longer. Actually, there's even a easier way to find the answer: f(n)=n^n + 2n + 1 in a certain sense behaves like a polynomial, since "If n is even then f(-1)=0 therefore f(n) is divisible by n + 1".
From: AP on 5 Apr 2010 11:47 On Mon, 05 Apr 2010 06:38:23 EDT, Christian Aebi <christian.aebi(a)edu.ge.ch> wrote: >Many thanks to all those who supplied an answer (even 12 years ago!). >No Kent, it wasn't homework and I should have thought a little longer. >Actually, there's even a easier way to find the answer: > f(n)=n^n + 2n + 1 in a certain sense behaves like a polynomial, since "If n is even then f(-1)=0 therefore f(n) is divisible by n + 1". or f(n)=(n+1-1)^n+2n+1=K(n+1)+(-1)^n+2n+1 and if n even, f(n)=K(n+1)+2(n+1)
From: Man. on 6 Apr 2010 00:17 On Apr 5, 3:38 am, Christian Aebi <christian.a...(a)edu.ge.ch> wrote: > Many thanks to all those who supplied an answer (even 12 years ago!). > No Kent, it wasn't homework and I should have thought a little longer. > Actually, there's even a easier way to find the answer: > f(n)=n^n + 2n + 1 in a certain sense behaves like a polynomial, since "If n is even then f(-1)=0 therefore f(n) is divisible by n + 1". Re: iscomposite( n^n + 2*n + 1) always ? No: isprime( 2^2 + 2*2 +1) = 7. Musatov
From: Man. on 6 Apr 2010 00:18 On Apr 5, 9:17 pm, "Man." <marty.musa...(a)gmail.com> wrote: > On Apr 5, 3:38 am, Christian Aebi <christian.a...(a)edu.ge.ch> wrote: > > > Many thanks to all those who supplied an answer (even 12 years ago!). > > No Kent, it wasn't homework and I should have thought a little longer. > > Actually, there's even a easier way to find the answer: > > f(n)=n^n + 2n + 1 in a certain sense behaves like a polynomial, since "If n is even then f(-1)=0 therefore f(n) is divisible by n + 1". > > Re: iscomposite( n^n + 2*n + 1) always ? > Yes: iscomposite( 2^2 + 2*2 +1) = 9. > Musatov Just checking soemthing.... :)
From: Rob Johnson on 6 Apr 2010 02:07
In article <1189307319.503074.1270463934038.JavaMail.root(a)gallium.mathforum.org>, Christian Aebi <christian.aebi(a)edu.ge.ch> wrote: >Many thanks to all those who supplied an answer (even 12 years ago!). >No Kent, it wasn't homework and I should have thought a little longer. >Actually, there's even a easier way to find the answer: > f(n)=n^n + 2n + 1 in a certain sense behaves like a polynomial, since >"If n is even then f(-1)=0 therefore f(n) is divisible by n + 1". Perhaps you can follow what you have written, and if I change some variables so that all variables are not "n", I can twist some sense into it, but it may be hard for some to read what you have written. At the risk of being repetitious (after 12 years), here is why n^n + 2n + 1 is always composite. If n is odd, n^n + 2n + 1 is even and greater than 2; therefore, it is composite. x - 1 divides x^k - 1 for any non-negative integer k. If k is odd, -(x + 1) = (-x) - 1 divides (-x)^k - 1 = -(x^k + 1); thus, x + 1 divides x^k + 1. If n is even, n-1 is odd, so n + 1 divides n^{n-1} + 1; thus, n+1 divides n (n^{n-1} + 1) + (n+1) = n^n + 2n + 1. Therefore, if n is even, n^n + 2n + 1 is divisible by, and greater than, n+1; therefore, it is composite. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |