More fun with the Prince of Maths
If we write 2 0 3 + 2 5 9 i as a product of Gaussian Primes , how many factors will there be?
The answer is 5.
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3 solutions
Yes! (+1) The beauty of the problem is that we don't actually have to find the prime factors to count them. We know that ( 2 0 3 + 2 5 9 i ) ( 2 0 3 − 2 5 9 i ) = 1 0 8 2 9 0 has 10 Gaussian prime factors, as you say, so 2 0 3 + 2 5 9 i and 2 0 3 − 2 5 9 i have 5 Gaussian prime factors each.
The norm is 2 0 3 2 + 2 5 9 2 = 1 0 8 2 9 0 . Factor out squares, twos, and prime factors of the form 4 k + 1 ; that gives 1 0 8 2 9 0 = 7 2 ⋅ 2 ⋅ 5 ⋅ 1 3 ⋅ 1 7 , each of which corresponds to a Gaussian prime factor. Thus there are five.
Yes! (+1) You are counting the prime factors without going through the trouble of finding them, as I do in my comment to Comrade Cheong's solution.
2 0 3 + 2 5 9 i = 7 ( 1 + i ) ( 2 + i ) ( 4 + i ) ( 3 − 2 i ) is a product of 5 Gaussian primes.
@Mark Hennings @Otto Bretscher I apologize for the confusion. An incorrect definition was indeed edited in by the moderators, which Mark objected to and Otto has since removed.
We know that the norm of a Gaussian integer is multiplicative, therefore the norm of 2 0 3 + 2 5 9 i must equal to the product of norms of its factors.
N ( 2 0 3 + 2 5 9 i ) = 1 0 8 2 9 0 = 2 × 5 × 7 2 × 1 3 × 1 7 ,
⇒ Norm 2 5 4 9 1 3 1 7 p ∈ N p = 2 5 = 4 k + 1 7 = 4 k + 3 1 3 = 4 k + 1 1 7 = 4 k + 1 Gaussian prime 1 + i 2 ± i 7 3 ± 2 i 4 ± i
2 0 3 + 2 5 9 i = 7 ( 2 9 + 3 7 i ) = 7 ( 1 + i ) ( 3 3 + 4 i ) = 7 ( 1 + i ) ( 2 + i ) ( 1 4 − 5 i ) = 7 ( 1 + i ) ( 2 + i ) ( 3 − 2 i ) ( 4 + i )
Therefore, there are 5 Gaussian prime factors.