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I have the following question:

Show that if $x,y\in\mathbb{Z}$ are positive and $z=x\pm yi$ and $x^2+y^2\equiv1\mod{4}$ then $z$ is irreducible in $\mathbb{Z}[i]$.

I know that a prime can be expressed as the sum of two squares iff the prime is congruent to $1$ modulo $4$, but I'm not sure how to interpret the sum of square being congruent to $1$ mod $4$.

I'm assuming that this requires a proof by contradiction, where you take $z=ab$ and derive the contradiction by using $N(ab)=N(a)N(b)=N(z)$, but I can't seem to find it.

高田航
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  • If $x^2+y^2\equiv 1\pmod 4$, then $(3x)^2+(3y)^2\equiv 1\pmod 4$ as well, but $z$ and $3z$ can't both be irreducible... – Mastrem Oct 22 '19 at 14:33

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Suppose that $x$ and $y$ are coprime. Then $$\Bbb Z[i]/(x\pm iy)\cong \Bbb Z/(x^2+y^2)\Bbb Z$$ is a finite integral domain, hence a field. Thus $(x\pm iy)$ is a maximal ideal and $x\pm iy$ is irreducible. Otherwise, $\operatorname{gcd}(x,y)>1$ and $x\pm iy$ is not prime, hence not irreducible.

Your claim is thus not correct. Take $(x,y)=(6,3)$. Then $$N(6+3i)=45\equiv 1\bmod 4,$$ but $6+3i=3(2+i)$ is not irreducible, because $2+i$ and $3$ are not units.

References:

Prove that $\mathbb{Z}[i]/\langle a+ib\rangle$ is isomorphic to $\mathbb{Z}/\langle a^2+b^2\rangle$ where $a, b$ are relatively prime

Quotient ring of Gaussian integers

Why does the ideal $(a+bi)$ have index $a^2+b^2$ in $\mathbb{Z}[i]$?

Quotient ring of Gaussian integers $\mathbb{Z}[i]/(a+bi)$ when $a$ and $b$ are NOT coprime

Dietrich Burde
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