This is part of a much larger problem.
I need to prove that:
$\displaystyle \forall n\in\mathbb{Z}, n^2 \neq (4k+3),\forall k\in\mathbb{Z}$
Please help me in any way you can. Thanks.
This is part of a much larger problem.
I need to prove that:
$\displaystyle \forall n\in\mathbb{Z}, n^2 \neq (4k+3),\forall k\in\mathbb{Z}$
Please help me in any way you can. Thanks.
Hello, miatchguy!
Prove that: .$\displaystyle \forall n\in\mathbb{Z},\: n^2 \neq (4k+3),\:\forall k\in\mathbb{Z}$
$\displaystyle \,n$ must be either even or odd: .$\displaystyle \begin{Bmatrix} n \:=\: 2m \\ \text{or} \\ n \:=\: 2m+1 \end{Bmatrix}$
If $\displaystyle \,n$ is even: .$\displaystyle n^2 \;=\;(2m)^2 \;=\;4m^2$
. . $\displaystyle \,n^2$ is a multiple of 4.
If $\displaystyle \,n$ is odd: .$\displaystyle n^2 \;=\;(2m+1)^2 \;=\;4m^2 + 4m + 1 \;=\;4(m^2+m)+1$
. . $\displaystyle \,n^2$ is one more than a multiple of 4.
Therefore, $\displaystyle \,n^2$ is never three more than a multiple of 4.