Strange Integers
Fundamental Theorem of Arithmetic
3 is not prime in Z[√3]. Give it a little thought, and the result is not at all surprising. This is what √3 was invented for  √3 times √3 is 3. 5 is a prime number. But what about 2? Is it a prime? No, it is not. We can write
(1)  2 = (5 + 3√3)(5 + 3√3), 
where none of the factors is a unity (check their Nvalue.)
In fact, once we found one factorization, we can easily obtain many more. Recollect that U = 2 + √3 as well as U' = 2  √3 are unities. Multiply the first factor in (1) by U and the second by U'. Apply the same procedure to the result:
(2) 

Note that the Nvalue of any factor is 2. This implies that each of them is prime.
So we learn not only that, in Z[√3], 2 is not prime, but also that its factorization into prime factors is not unique. Indeed, in the presence of an infinite number of unities, no number may have a unique factorization. However, in (2) factors differ by a multiple of unity. Such primes are called associates. For example, 5 + 3√3 and 71 + 41√3 are associates, for
By the end of this page I plan to prove the following variant of the Fundamental Theorem of Arithmetic:
In Z[√3], factorization of a number into a product of primes is unique up to the order of associated primes.
First let's establish an analogue of the Euclidean Algorithm:
(3) 
For any given pair A and B from Z[√3], B≠0, there exist in Z[√3] numbers Q and R such that. A = BQ + R, and N(R) < N(B) 
Let A/B = x + y√3, where x and y are rational. Any rational number p lies somewhere between [p] and [p]+1, where [p] is the largest integer smaller than p. Therefore, any rational number is at the distance of at most ½ from an integer. Let r and s be integers such that
Let Q = r + s√3 and R = B(x  r) + B(y  s)√3. Then, of course,
N(R) = N(B)((x  r)²  3(y  s)²). By the defintion,
(4)  (x  r)²  3(y  s)² ≤ x  r² + 3y  s² ≤ (½)² + 3(½)² = 1. 
Equality is only achieved when both x  r and y  s equal ½. But in this case,
(x  r)²  3(y  s)² = ½²  3½² = ½ < 1
Therefore always (x  r)²  3(y  s)² < 1, and we have
(3) is a very powerful result. Now we may start an analogue of Euclid's Algorithm with confidence, assured that it will terminate in a finite number of steps. When it stops, we shall have found a number D with the property that
 DA and DB, and
 For any C such that CA and CB, we also have CD.
This D is naturally called the greatest common divisor of A and B, and is denoted as
AS + BT = gcd(A, B)
The greatest common divisor is obviously defined up to a factor of unity. Two numbers are called coprime (mutually prime) iff their greatest common divisor is a unity. Since 1 associates with every unity,
AS + BT = 1,
for coprime A and B and some S and T; exactly as was the case with integers.
We are in a position to prove the most fundamental property of division:
(5)  Assume that a prime number P divides the product of two numbers A and B: PAB. Then either PA or PB. 
In other words, if, for instance, P∤A, then it must be that PB. As a prime, the only factors of P are its associates and the unities. If any of the associates divided A, so would P, but, by our assumption, this is not the case. Therefore, gcd(P,A) = 1. We thus conclude that there exist two numbers S and T such that PS + AT = 1. Multiply by B: PBS + (AB)T = B. Now, PPBS and PAB. Whence PB.
Uniqueness of the prime factorization follows immediately from (5). Let A = P_{1}P_{2}...P_{n} = Q_{1}Q_{2}...Q_{m}, where all P's and Q's are prime. P_{1} divides Q_{1}Q_{2}...Q_{m}. Therefore, by (5), it divides one of the Q's. Both being prime, they are associates of each other and their ratio is a unity. Assume P_{1}Q_{1} and Q_{1}/P_{1} = U. Replace Q_{2} with UQ_{2}. This will not affect the primality of Q_{2}. We can chip off factors from one side or another until we get, for example, Q_{1}Q_{2}...Q_{k} = 1, all other factors from the two sides having been matched. The remaining Q's are all unities, hence, not prime. Whence
Remark
6 = 3·2 = 3(31) = 3²  3 = (3 + √3)(3  √3). Since 26, we also have that 2(3 + √3)(3  √3). However, 2∤(3 + √3) and 2∤(3  √3), for, say, (3 + √3)/2 = 3/2 + √3/2 which does not belong to Z[√3] at all. Combined with (4), we get another proof that 2 is not prime in Z[√3].
Question
Does or does not the identity 3·2 = (3 + √3)(3  √3) contradict the fact that the prime factorization in Z[√3] is unique?
Reference
 H. M. Stark, An Introduction to Number Theory, The MIT Press, 1995, Ninth printing.
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