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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?

N. F. Taussig
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user476504
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    You haven't written all details about what you proved about the square of an integer being either $0\pmod{4}$ or $1\pmod{4}$ (in particular, you only wrote about what happens for $x$ even, not yet about $x$ odd), but yes that is a correct approach. – JMoravitz Sep 01 '17 at 17:08
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    All of the work here boils down to showing that the square of an integer is either $0$ or $1 \pmod{4}$. If there were a solution, then modding out both sides of that equation gives $x^2 \equiv 3 \pmod{4}$. Uh oh. – Kaj Hansen Sep 01 '17 at 17:09
  • Thanks! If I could accept both of your comments as answers I would! – user476504 Sep 01 '17 at 17:10

2 Answers2

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$$x^2=\underbrace{4y}_{even}+\underbrace{3}_{odd} \to x \text { must be odd }$$ 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

Khosrotash
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$x$ 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 $$4y+3\equiv 3\pmod{4}$$

dromastyx
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