The equation $x^2+3y^2=m$ has no solutions when $m\equiv 2\bmod 4$

Question

For every integer $m$ such that $2$ divides $m$, and $4$ doesn't divide $m$, there are no integers, $x$ and $y$ that satisfy $x^2 + 3y^2 = m$.

Use a contradiction (assume the negation is true)

Is my negation of the statement correct? My negation is: There exists an integer m such that $2$ doesn't divide m OR 4 does divide m, for all integers $x,y$, there are NONE that satisfy $x^2 + 3y^2 = m$.

By divides I mean: for example 2 divides 6 because 2 is a factor of 6. similarly by 2 divides m, i mean 2 is a factor of m.

Answer 1

We have $m\equiv2\pmod4$

Any integer $a\equiv0,1,2,3\pmod4\implies a^2\equiv0,1\pmod4$

then we can establish $x^2+3y^2\not\equiv2\pmod4$

Answer 2

If $x^2+3y^2=4a+2\iff(x+y)(x-y)=2(2a+1-2y)$

Now for integers $x,y; x\pm y$ have the same parity as $x+y-(x-y)$ is even

If one is odd, the other & consequently the product must be odd

If one is even, the other will be even & consequently the product must be divisible by $4$

Answer 3

Check the options of $x^2$ and $3y^2$ $\text{mod }4$.

$$\begin{array}{|cc|c|cc|} x & x^2 & & y & 3y^2\\ 0 & 0 & & 0 & 0 \\ 1 & 1 & & 1 & 3 \\ 2 & 0 & & 2 & 0 \\ 3 & 1 & & 3 & 3 \\ \end{array}$$

Therefore $x^2 + 3y^2 \equiv 0,1,\text{ or }3 \pmod{4}$

But $m \equiv 2 \pmod 4$. Therefore there are no solutions for any such $m$.