Prouve that number is not prime number :
(x is whole number no zero ) .
Hello, dhiab!
Prove that $\displaystyle N \;=\;2^{2n+1}+1$ is not a prime number, for any positive integer $\displaystyle n.$
$\displaystyle N \;=\;2^{\text{odd power}} + 1$ is always factorable.
. . $\displaystyle 2^{2n+1} + 1 \;=\;(2 + 1)\left(2^{2n} - 2^{2n-1} + 2^{2n-2} + \hdots + (\text{-}1)^{n+1}\right) $
Therefore, $\displaystyle N$ is always a multiple of 3.
[quote=Soroban;326807]Hello, dhiab!
$\displaystyle N \;=\;2^{\text{odd power}} + 1$ is always factorable.
. . $\displaystyle 2^{2n+1} + 1 \;=\;(2 + 1)\left(2^{2n} - 2^{2n-1} + 2^{2n-2} + \hdots + (\text{-}1)^{n+1}\right) $
Therefore, $\displaystyle N$ is always a multiple of 3.
[/quote
Thank you for this resolution ?This my resolution (Theorem of induction)
I prouve that : :
- for : I have .
-Now . I PROUVE FOR :
He3llo, again, dhiab!
This is my resolution (induction)
I prove that: .$\displaystyle \forall x \in I^+,\;2^{2x+1} + 1 \:=\:3k$
For $\displaystyle x = 1$, I have: .$\displaystyle 2^{2(1)+1} + 1 \:=\:2^3+1 \:=\:9 \:=\:3\cdot3$
Now: .$\displaystyle 2^{2x+1} + 1 \:=\:3k\:\text{ for some integer }k.$
Prove for $\displaystyle x+1$:
$\displaystyle 2^{2(x+1)+1}+1 \;=\;2^{2x+3} + 1 \;=\;4\cdot2^{2x+1} + 1 \;=\;3\cdot2^{2x+1} + \underbrace{(2^{2x+1}+1)}_{\text{This is }3k} $
Therefore: .$\displaystyle 2^{2(x+1)} + 1 \;=\;3\cdot2^{2k+1} + 3k \;=\;3\bigg[2^{2k+1} + k\bigg] \;=\;3m\:\text{ for some integer }m.$
Your inductive proof is correct!