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.

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.

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 \(x=1\).

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 \(z=ac+bd\).

It suffices to show that \(p\) is composite in \(\mathbb{Z}[\omega]\). Indeed, if there is a prime element \(z=a+b\omega\in\mathbb{Z}[\omega]\) (\(a,b\in\mathbb{Z}\)) that divides \(p\), then also \(\overline{z}\mid\overline{p}=p\). Note that \(z\) and \(\overline{z}\) are coprime; otherwise \(z\mid\overline{z}\), so there exists a unit element \(\epsilon\) with \(\overline{z}=\epsilon z\), and hence \(z\sim (1-\omega)\mid 3\), which is false. Therefore \(a^2-ab+b^2=z\overline{z}\mid p\), which implies \(a^2-ab+b^2=p\).

Thus we need to prove that \(p\) is composite in \(\mathbb{Z}[\omega]\). It follows from the condition on \(p\) that -3 is a quadratic residue modulo \(p\), so there exist \(m,n\in\mathbb{Z}\) which are not divisible by \(p\) such that \(p\mid(2m-n)^2+3n^2=4(m^2-mn+n^2)\), i.e. \(p\mid(m-n\omega) \overline{m-n\omega}\). However, \(p\) does not divide any of the elements \((m-n\omega),\overline{m-n\omega}\), so it must be composite.

Suppose that \(x,y,z\) are nonzero elements of \(\mathbb{Z}[\omega]\) that satisfy \((\ast)\). We can assume w.l.o.g. that \(x,y,z\) are pairwise coprime.

Consider the number \(\rho=1-\omega\). It is prime, as its norm equals \((1-\omega)(1-\omega^2)=3\). We observe that \(\overline{\rho}=1-\omega^2=(1-\omega)(1+\omega)\sim\rho\); hence \(\alpha\in\mathbb{Z}[\omega]\) is divisible by \(\rho\) if and only if so is \(\overline{alpha}\). Each element in \(\mathbb{Z}[\omega]\) is congruent to \(-1,0\) or 1 (mod \(\rho\)): indeed, \(a+b\omega\equiv a+b=3q+r\equiv r\) (mod \(\rho\)) for some \(q\in\mathbb{Z}\) and \(r\in\{-1,0,1\}\).

The importance of number \(\rho\) lies in the following property: \[\alpha\equiv\pm1\; (\mbox{mod }\rho)\; (\alpha\in\mathbb{Z}[\omega])\;\;\mbox{implies}\;\; \alpha^3\equiv\pm1\;(\mbox{mod }\rho^4). \quad\quad\quad\quad\quad(2)\] Indeed, if \(\alpha=\pm1+\beta\rho\), we have \(a^3\mp1=(a\mp1)(a\mp\omega) (a\mp\omega^2)=\rho^3\beta(\beta\pm1)(\beta\pm(1+\omega))\), where the elements \(b\), \(b\pm1\), \(b\pm(1+\omega)\) leave three distinct remainders modulo \(\rho\), implying that one of them is also divisible by \(\rho\), thus justifying our claim.

Among the numbers \(x,y,z\), (exactly) one must be divisible by \(\rho\): otherwise, by (2), \(x^3,y^3,z^3\) would be congruent to \(\pm1\) (mod \(\rho^4\)), which would imply one of the false congruences \(0\equiv\pm1\), \(\pm1\equiv\pm2\) (mod \(\rho^4\)). We assume w.l.o.g. that \(\rho\mid z\). Moreover, (2) also gives us that \(\rho^2\mid z\).

Let \(k\geq2\) be the smallest natural number for which there exists a solution to \((\ast)\) with \((x,y,z)=1\) and \(\rho^k\mid z\), \(\rho^{k+1}\nmid z\). Consider this solution \((x,y,z)\).

The factors \(x+y\), \(\omega x+\omega^2y\), \(\omega^2x+\omega y\) from (1) are congruent modulo \(\rho\) and have the sum 0. It follows from \(\rho\mid z\) that each of them is divisible by \(\rho\) and that \(\rho\) is their greatest common divisor. Let \[x+y=A\rho,\;\;\;\omega x+\omega^2y=B\rho,\;\;\;\omega^2x+\omega y=C\rho,\] where \(A,B,C\in\mathbb{Z}[\omega]\) are pairwise coprime and \(A+B+C=0\). The product \(ABC\) is a perfect cube (equal to \((z/\rho)^3\)), and hence each of \(A,B,C\) is adjoint to a cube: \[A=\alpha\zeta^3,\;\;\;B=\beta\eta^3,\;\;\;C=\gamma\xi^3\] for some pairwise coprime \(\zeta,\eta,\xi\in\mathbb{Z}[\omega]\) and units \(\alpha,\beta,\gamma\). Therefore, \[\alpha\zeta^3+\beta\eta^3 +\gamma\xi^3=0.\quad\quad\quad\quad\quad(3)\] Since \(\alpha\beta\gamma\) is a unit and a perfect cube, we have \(\alpha\beta\gamma=\pm1\). Furthermore, \(ABC=(z/\rho)^3\) is divisible by \(\rho\) (since \(\rho^2\mid z\)), so (exactly) one of the numbers \(\zeta,\eta,\xi\), say \(\xi\), is divisible by \(\rho\). In fact, \(\xi^3\) divides \(ABC\) which is divisible by \(\rho^{3k-3}\) and not by \(\rho^{3k-2}\), so \(\rho^{k-1}\) is the greatest power of \(\rho\) that divides \(\xi\). The numbers \(\zeta\) and \(\eta\) are not divisible by \(\rho\); consequently, \(\zeta^3\) and \(\eta^3\) are congruent to \(\pm1\) modulo \(\rho^4\). Thus the equality \(A+B+C=0\) gives us \(\alpha\pm\beta\equiv0\) (mod \(\rho^4\)), therefore \(\beta=\pm \alpha\); now \(\alpha\beta\gamma=\pm1\) implies \(\gamma=\pm\alpha\).

Canceling \(\alpha\) in equation (3) yields \(\zeta^3\pm\eta^3\pm\xi^3=0\), which gives us another nontrivial solution to \((\ast)\) with \((\zeta,\eta,\xi)=1\). However, in this solution we have \(\rho^{k-1}\mid\xi\) and \(\rho^k\nmid\xi\), which contradicts the choice of \(k\).