# Arithmetics in Gaussian Integers $$\mathbb Z[i]$$

We have already seen that the norm of element $$a+bi\in\mathbb{Z}[i]$$ ($$a,b\in\mathbb{Z}$$) is $$N(a+bi)=a^2+b^2$$ and the units are $$\pm1$$ and $$\pm i$$. Therefore, all divisors of a prime element $$\pi\in\mathbb{Z}[i]$$ are $$\pm1,\pm i,\pm\pi,\pm i\pi$$.

Theorem 6 The Fundamental Theorem of Arithmetic (FTA) holds in the set of Gaussian integers $$\mathbb{Z}[i]$$.

Based on theorem 5, it is enough to show that for all $$a,b\in\mathbb{Z}[i]$$ with $$b\neq0$$ there exists an element $$p\in\mathbb{Z}[i]$$ such that $$N(a-pb) < N(b)$$.

Let $$\sigma,\tau\in\mathbb{R}$$ be such that $$a/b=\sigma+\tau i$$, and let $$s,t\in\mathbb{Z}$$ be such that $$|\sigma-s|\leq 1/2$$ and $$|\tau-t|\leq1/2$$. Setting $$p=s+ti$$ yields $$a-pb=(\sigma+\tau i)b-pb=[(\sigma-s)+(\tau-t)i]b$$, which implies $N(a-pb)= N[(\sigma-s)+(\tau-t)i]N(b)=[(\sigma-s)^2+(\tau-t)^2]N(b)\leq N(b)/2 < N(b).$ This proves the theorem.

The following proposition describes all prime elements in the set of Gaussian integers.

Theorem 7 An element $$x\in\mathbb{Z}[i]$$ is prime if and only if $$N(x)$$ is a prime or $$|x|$$ is a prime integer of the form $$4k-1$$, $$k\in\mathbb{N}$$.

Consider an arbitrary prime $$x=a+bi \in\mathbb{Z}[i]$$ ($$a,b\in \mathbb{Z}$$). Element $$\overline{x}$$ is prime also (indeed, if $$\overline{x}=yz$$, then $$x=\overline{y}\: \overline{z}$$), so $$N(x)$$ factorizes into primes as $$x\overline{x}$$.

Suppose that $$N(x)$$ is composite, $$N(x)=mn$$ for some two natural numbers $$m,n > 1$$. It follows from $$x\overline{x}=mn$$ and the FTA that $$x\sim m$$ or $$x\sim n$$, so we may suppose w.l.o.g. that $$x$$ is a prime integer. If $$x=2$$ or $$x\equiv1$$ (mod 4), then there exist integers $$a,b\in\mathbb{Z}$$ such that $$N(a+bi)=(a+bi)(a-bi)=a^2+b^2=x$$; hence $$x$$ is composite in $$\mathbb{Z}[i]$$. On the other hand, if $$x$$ is a prime integer with $$x\equiv3$$ {\rm(mod 4)}, then $$x$$ is also prime in $$\mathbb{Z}[i]$$. Indeed, if $$x=uv$$ for some nonunit elements $$u,v\in\mathbb{Z}[i]$$, then $$x^2=N(x)=N(u)N(v)$$ implies $$N(u)=N(v)=x$$ which is impossible. This proves our claim.

Problem 3 Solve the equation $$x^5-1=y^2$$ in integers.

Rewrite the equation in the form $$x^5=(y+i)(y-i)$$. Note that $$x$$ is not even, as otherwise $$y^2\equiv-1$$ (mod 4). Thus $$y$$ is even and consequently the elements $$y+i$$ and $$y-i$$ are coprime in $$\mathbb{Z}[i]$$. Since $$(y+i)(y-i)$$ is a fifth power, it follows that $$y+i$$ and $$y-i$$ are both fifth powers. Let $$a,b\in \mathbb{Z}$$ be such that $$y+i=(a+bi)^5=a(a^4-10a^2b^2+5b^4)+b(5a^4-10a^2b^2+b^4)i$$. It holds that $$b(5a^4-10a^2b^2+b^4)=1$$, and therefore $$b=\pm1$$. It is easily seen that we must have $$y=0$$ and $$x=1$$.

Problem 4 Suppose that $$x,y$$ and $$z$$ are natural numbers satisfying $$xy=z^2+1$$. Prove that there exist integers $$a,b,c,d$$ such that $$x=a^2+b^2$$, $$y=c^2+d^2$$ and $$z=ac+bd$$.

We use the following important fact: If $$m,n,p,q$$ are elements of a unique factorization domain $$K$$ (in this case, $$K=\mathbb{Z}[i]$$) satisfying $$mn=pq$$, then there exist $$u_1,u_2,v_1,v_2\in K$$ such that $$m=u_1v_1$$, $$n=u_2v_2$$, $$p=u_1v_2$$, $$q=u_2v_1$$. The proof of this fact is the same as in the case of $$m,n,p,q\in\mathbb{Z}$$ and follows directly from the factorizations of $$m,n,p,q$$ into primes.

Since $$xy=z^2+1=(z+i)(z-i)$$, the above fact gives us $\begin{array}{cccccccc}x=u_1v_1&(1), &y=u_2v_2&(2),&z+i=u_1v_2&(3),&z-i=u_2v_1&(4)\end{array}$ for some $$u_1,u_2,v_1,v_2\in\mathbb{Z}[i]$$. The numbers $$x,y$$ are real, and therefore $$v_1=q_1\overline{u_1}$$, $$v_2=q_2\overline{u_2}$$ for some rational numbers $$q_1,q_2$$. From (3) and (4) we easily conclude that $$q_1=q_2=1$$. Now putting $$u_1=a+bi$$, $$u_2=c+di$$ yields $$x=u_1\overline{u_1}=a^2+b^2$$, $$y=c^2+d^2$$ and $$z=ac+bd$$.