TITLE: Direct Proof - Discrete Mathematics
QUESTION [5 upvotes]: I am given
Prove that there are no integer solutions to the equation
$$x^2=4y+3$$
I started off by proving the square of the integer is either $0 \pmod{4}$ or $1 \pmod{4}$. If $x$ is even then $x=2k$ for some integer $k$. Then $x^2=(2k)^2=4k^2$.
Will this satisfy the question?

REPLY [4 votes]: $$x^2=\underbrace{4y}_{even}+\underbrace{3}_{odd} \to x \text { must be odd }$$
take x odd like 2k+1
  so 
$$x^2=(2K+1)^2=4k^2+4k+1=4\underbrace{k(k+1)}_{even}+1\\\to x^2=8q+1$$ now check in first equation 
$$x^2=4y+3\\8q+1=4y+3\\8q-4y=2\\\div 4 \\\underbrace{2q-y}_{\in \mathbb{Z}}=\frac 12$$and it is impossible

REPLY [3 votes]: $x$ must be odd (since rhs is odd), i.e $x=2k+1$ for $k\in\mathbb{Z}$. But then lhs is
$$x^2=(2k+1)^2=4(k^2+k)+1\equiv 1\pmod{4}$$
but rhs
$$4y+3\equiv 3\pmod{4}$$