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.

- Oct 8th 2010, 12:50 AMmiatchguyI need to prove that a term of a certain form cannot be a perfect square.
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. - Oct 8th 2010, 01:38 AMaman_cc
Hint - Take n = 4m + 1 or 4m+3 ? What do you see? You don't need to consider even 'n'

- Oct 8th 2010, 05:37 AMSoroban
Hello, miatchguy!

Quote:

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, $\displaystyle \,n^2$ isthree more than a multiple of 4.*never*