How can I show that $a^2 + b^2 \equiv 3(\mod 4)$ has no integer solutions?

Question

I want to show that $a^2 + b^2 \equiv 3(\mod4)$ has no solutions. I know this equation will have solutions if gcd$(4, a^2)$ divides $(3-b^2)$; I need to show that this isn't true. WolframAlpha confirms that the equation doesn't have solutions, but I have no idea how to show this. Help would be much appreciated.

https://i.stack.imgur.com/3JzNX.png

Answer 1

Notice that $3=1+2=3+0$. But the only values that be remainders of perfect squares when divided by $4$ are $0$, $1$$^*$. So it is not possible to have $\bmod 4: x^2+y^2\equiv 3$.

However you can find integer solutions to $$\bmod 4: x^2+y^2 \equiv 2,1,0$$

Proof of $^*$: Any integer can be of form $4k$, $4k+1$, $4k+2$ and $4k+3$. $$\bmod 4: \\ (4k)^2 \equiv 0 \\ (4k+1)^2 \equiv 1 \\ (4k+2)^2 = 16k^2+4+16k \equiv 0 \\ (4k+3)^2 = 16k^2+24k+9 \equiv 1$$

Answer 2

Modulo $4$ you're only working with the integers $0,1,2,3\in\Bbb Z_4$. One way to study the expression $a^2+b^2$ is to check what the squares look like in $\Bbb Z_4$.

$$\begin{array} {|r|r|}\hline x & x^2 \\ \hline 0 & 0 \\ \hline 1 & 1 \\ \hline 2 & 0 \\ \hline 3 & 1 \\ \hline \end{array}$$ So any square in $\Bbb Z_4$ is either $x^2\equiv_4 0$ or $x^2\equiv_4 1$. Now let's look at the possible values of $a^2+b^2$, and see if this can be made to be $3$.

$$\begin{array} {|r|r|}\hline +_4 & 0 & 1 \\ \hline 0 & 0 & 1 \\ \hline 1 & 1 & 2 \\ \hline \end{array}$$ As we can see it is not possible to get $3$, so $$a^2+b^2\equiv_4 3$$ has no solution.

Answer 3

Here's another way to do it.

Suppose $a^2 + b^2 \equiv_4 3$ .

Then $a^2 + b^2 \equiv_2 3 $ .

Then $a + b \equiv_2 1$ .

Suppose without loss of generality that $a = 2g+1$ is odd and $b = 2h$ is even.

Consider $(2g+1)^2 + (2h)^2 \mod 4$ .

This is equivalent to $4g^2 + 4g + 1 + 4h^2 \mod 4$ .

Which is just $1 \mod 4$ .

Thus $a^2 + b^2 \equiv_4 1$, contradicting our hypothesis.