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Σωτήριος Δ. Χασάπης - www.arithmoi.gr 2000-2020 - Arithmetics in Gaussian Integers - Page 2

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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 \(z=ac+bd\).

 

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