How to prove that 2 is not a Gaussian prime?

Question

To prove an algebraic number is not a Gaussian prime, need find factors of it. Let us assume, $(a + bi)(x +yi) = 2, \exists a,b,x,y \in \mathbb{Z}$; so $(ax -by) + i(ay + bx) =2$.

As imaginary part is null in $2$, so $ax - by =2, ay + bx =0$.

Now, how to reach from here, the fact that $a=x=1, b=i, y=-i$.

I read in here in a comment by @StevenStadnicki, that an easier way is to multiply both sides of $(a + bi)(x +yi) = 2$ by the complex conjugates, but is not clear, as my below attempt shows. $$ \begin{align} & ax - by =2, ay + bx =0 & \ \\ \implies & axy - by^2 =2y, axy + bx^2 =0 \text{ }{ -(i)} & \ \\ \implies & b = 0 \end{align} $$

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By the answer given below, my attempt is stated below:

Two set of values are possible, one in which $(a^2 + b^2) = 4, (x^2 + y^2) = 1$, second in which $(a^2 + b^2) = 2, (x^2 + y^2) = 2$.

Possible combinations for each of the two cases are stated below:

Case (i): $(a^2 + b^2) = 4, (x^2 + y^2) = 1$
$\implies a= \pm 2, b = 0, x = \pm 1, y = 0$, or $a= \pm 2, b = 0, x = 0, y = \pm 1$.
But, as $ax -by =2, ay + bx = 0$.
But being algebraic number, need have non-zero imaginary part, hence no algebraic factors of $2$.

Case (ii) $(a^2 + b^2) = 2, (x^2 + y^2) = 2$
$\implies$ possible values are : $ a= \pm 1, b = \pm 1, x = \pm 1, y = \pm 1$.

But, as : $ax -by =2, ay + bx = 0$, so the valid combinations are:
1. $a=1, b=-1, x=1, y=1$;
2. $a=1, b=1, x=1, y=-1$,
3. $a=-1, b=1, x=-1, y=-1$.
Here, combination 3 is same as 1, so can be ignored also valid.

So, the solution has values of $a =1,b=1, x = 1, y = -1$.

If there is a redundancy or unnecessary steps, please tell.

Answer 1

Let $(a+ib)(c+id)=2$.

There is obviously no solution with real integers, so $b$ and $d$ are both nonzero (if only one is zero, the product is complex or 0, so it also fails). Also, since the product is real, their arguments are opposite: they are multiples of a conjugate pair of gaussian integers.

So it's really $k(a+ib)(a-ib)=2$

Hence

$$k(a^2+b^2)=2$$

So $k$ must be positive. And $k,a,b$ are integers. If $k=2$, it fails because then $b=\pm1$ and $a=0$ (remember $b\ne0$), and the factorization if $2i\times -i=2$ or $-2i\times i=2$, and this proves nothing as you have just multiplied $2$ by a unit.

Hence let $k=1$ and $a^2+b^2=2$, but then $a^2=b^2=1$. But the sign of $a$ is useless, because $(-a+ib)(-a-ib)=(a+ib)(a-ib)$. Hence, you can pick $a=1$, and the solution has to be

$$(1+i)(1-i)=2$$

Which you can check.

Of course, another solution is

$$(-1+i)(-1-i)=2$$

Answer 2

If $(a+bi)(x+yi)=2$, $(a-bi)(x-yi)=2$ and therefore $(a^2+b^2)(x^2+y^2)=4$. In how many ways can $4$ be written as the product of two integers, each of which is a sum of $2$ squares? Well, $4=4\times1=(2^2+0^2)\times(1^2+0^2)$ and $4=2\times2=(1^2+1^2)\times(1^2+1^2)$. Therefore, there are few possibilities for you to test.