Gaussian Integers II Let us summarise what we now about Gaussian integers so far: If a, b Z[ i], then N( ab) N( a) N( b). In particular if a b then N( a ) N( b). Let z Z[i]. If N( z ) is an integer prime, then z is a Gaussian prime. A prime p Z is a Gaussian prime if and only if p y has no solution. Fundamental Theorem of Arithmetic Unique prime factorization Euclidean Division Theorem, Algorithm, and GCD algorithm GCD property: the gcd of two Gaussian integers is a linear combination of them. Prime Divisor Property If p is a Gaussian prime and p wz then p w or p z. We are going to use these properties to prove the following theorem: Fermat s Two Squares Theorem Let p be a prime. The equation p mod 4. y has a solution if and only if p = or p 1 We already now that y p has no solution when p 3 mod 4, by Theorem 3.30 in the boo: if y p then y 0 mod p, which by Theorem 3.30 implies 0 y mod p, and so and y are multiples of p. But it is then impossible that their squares add up to p in the integers. The main part of the two squares theorem is to show that if p 1 mod 4 then that equation can be solved. The ey is to figure out eactly which Gaussian integers are primes. 4. Fermat s Two Squares Theorem We already now that a composite integer cannot be a Gaussian prime. By Theorem.1, the primes p that are Gaussian primes are those with no solution to p y has no solution. Lemma.1 says that any Gaussian integer of prime norm is a Gaussian prime. We show these are the only other Gaussian primes: Theorem 4.1. The Gaussian primes are units times integer primes not of the form Gaussian integers of prime norm. y or
Proof. Let z denote the comple conjugate of a Gaussian integer z, so N ( zz. We have to show that if z is a Gaussian prime, then z is an integer prime or N ( is an integer prime. Suppose z is a Gaussian prime, but N ( is not an integer prime. Then N ( mn for some positive integers m > 1 and n > 1. Since z is a Gaussian prime, so is z [Eercise]. By the unique factorization theorem in the Gaussian integers, the unique prime factorization of N ( must be N ( zz. But N ( mn is another factorization, so m and n must be the Gaussian primes z and z up to multiplication by a unit. This means m and n must be an integer prime and its comple conjugate. But m > 1 and n > 1, so the only possibility is m n p for some integer prime p, and then z is p or a unit times p, as required. Lagrange s Lemma. If p 1 mod 4 is prime, then p n 1 for some integer n. Proof. Write p 4m 1. By Wilson s Theorem, and m i (m i 1) mod p, (4m)! (m )! 1 mod p. Putting n ( m)! we habe n 1 0 mod p. Fermat s Two Squares Theorem Let p be a prime. The equation p mod 4. y has a solution if and only if p = or p 1 Proof. Theorem 3.30 says if p y then p is not 3 mod 4 i.e. p = or p 1 mod 4. For the reverse implication, it is trivial for p = so we concentrate on p 1 mod 4. By Lagrange s Lemma, p n 1 for some integer n. Now n 1 ( n i )( n i ) and so by the prime divisor property, if p is a Gaussian prime, then p n i or p n i. These are both impossible [Eercise], so p can t be a Gaussian prime. By Theorem.1, that means p y has a solution. In fact one can show that there is only one solution to p y with y.
5. Pythagorean Triples We return to Diophantine equations. Specifically, let us study equations of the form y z where > 1 is an integer. This includes the eample of Pythagorean triples y z. At the beginning of the notes, we outlined the strategy: 1. Factor y in the Gaussian integers: y ( (.. Show that iy, iy are relatively prime 3. Use prime divisor property: (,( are both units times th powers. 4. Solve iy ( u iv) The third step is justified by the following lemma. Lemma 5.1 Let, y and z be Gaussian integers such that y z and suppose and y are relatively prime. Then for some unit u, and some Gaussian integers a and b, a and y b u. e e e Proof. If z z z r 1 1 r is the prime factorization of z, where the z i are distinct Gaussian e e e primes, the prime factorization of y is z z z r 1 1 r. By the prime divisor property, e z i e i y implies z i e i or z i i y for i = 1,...,r. If the first s of them divide, then we have z e1 e e 1 z z s s e i i ( z ) and similarly for y. The second step depends more on the equation. It is not true in general that a Gaussian integer and its comple conjugate are relatively prime, for instance i, i are not relatively prime. Lemma 5. Suppose y z for integers, y and z, where (, y, 1. Then iy, iy are relatively prime.
Proof. Let w be a GCD of iy, iy. We have to show w is a unit. We now that w ( w ( ( ( w iy w Since (, y, 1 and y z we have (, y) 1. So that means w. The prime factorization of is (1 i)(1 i) so w is a unit times 1 i or a unit times 1 i or a unit. But if w is a unit times 1 i or 1 i we have a problem: w y (1 i ) z i z z 4 z and mod 4 we have y 0. This implies 0 y mod 4, but that is impossible since (, y) 1. We conclude that w is a unit. With these two lemmas, we can now solve the Diophantine equation. Theorem 5.1 The positive integer solutions to of triples (, y, such that (, y, 1 and y z are given by all positive integer multiples y v or y v z z where u, ( u, v) 1, and eactly one of u and v is even. Proof. We conclude from Lemma 5.1 and 5. that there eists a unit t such that ( i y) ta,( b t for some Gaussian integers a and b. Suppose a U iv. Then the first equation gives ( i y) t( U iv) t( U V ) tiuv. Now t { 1, 1, i, i}. We consider each case.
If t = 1, then U V, y UV and we let u U, v V If t = -1, then V U, y UV and we let u V, v U If t = i, then y U V, UV and we let u U, v V If t = -i, then y V U, UV and we let u V, v U In all cases, we have acquired solutions of the required form and u > v since we are looing only at positive solutions. It remains to show eactly one of u and v is even, and (u,v) = 1. If u and v have a common factor, then it is also a common factor of and y and z, which contradicts (, y, 1. So (u,v) = 1. In particular, they can t both be even. If they are both odd, then, y and z would all be even, another contradiction to (, y, 1. So eactly one of u and v is even. Eample. Solve prime. By Lemma 5.1, 3 1 z. Well by Lemma 5. with y = 1, i, i are relatively i 3 3 ta, i b t for some unit t and some Gaussian integers imaginary parts we get a iv and b. If t 1, comparing 3 1 Im( u i v) (3u ). If t i comparing imaginary parts we get 1 Im i ( u iv) ( u 3v ). The only possibilities are ( u, v) (1,0 ) and ( u, v) (0,1 ). Both cases give 3 Re( u iv ) 3 0 and so the only solutions are 0, z 1.