n e N, and n = 5 (mod6). prove that n^2 + 2 is composite. Can someone give me a detailed, step by step proof of how to do this?? I'm completely confused when it comes to modular arithmetic
Hello,
$\displaystyle n\equiv 5(\bmod 6) \Rightarrow n^2 \equiv 5^2 (\bmod 6)\Rightarrow n^2 \equiv 1(\bmod 6)$
Hence $\displaystyle n^2+2\equiv 3(\bmod 6)$
So $\displaystyle \exists k\in\mathbb{N} ~:~ n^2+2=3+6k$
It should be obvious that 3 divides $\displaystyle n^2+2$. And since $\displaystyle n\equiv 5(\bmod 6)$, it follows that $\displaystyle n\geq 5$, and hence $\displaystyle n^2+2> 3$.
Therefore, $\displaystyle n^2+2$ is composite.
Hello, qtpipi!
Another approach . . .
Since $\displaystyle n \equiv 5\text{ (mod 6)}$, then $\displaystyle n \:=\:5 + 6k$ for some integer $\displaystyle k.$$\displaystyle n \in N\text{, and }n \equiv 5\text{ (mod 6)}$
Prove that $\displaystyle n^2 + 2$ is composite.
Let $\displaystyle A \:=\:n^2+2$
. . Then: .$\displaystyle A \:=\: (5+6k)^2 + 2 \;=\;25 + 60k + 36k^2 + 2 \;=\;36k^2 + 60x + 27$
$\displaystyle \text{Hence: }\;A \;=\;\underbrace{3(12k^2 + 20x + 9)}_{\text{a multiple of 3}}$
Therefore, $\displaystyle A$ is composite.