Do it like Gauss
A Gaussian Integer is a complex number of the form z = a + b i , where a and b are integers. How many Gaussian Integers z divide 10, in the sense that 1 0 = z × w for some Gaussian Integer w ?
The answer is 48.
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1 solution
Another way to do it is to prime factorize 10 as a gaussian integer, like this: 1 0 = − i ( 1 + i ) 2 ( 2 + i ) ( 2 − i ) Then you can see that there are 2 3 ( 3 ) pairs (by choosing factors for z ), and you may also multiply each in the pair by − 1 . This results in 4 8 total pairs.
Unfortunately, I was stupid and forgot the negative one part, and clicked show answer before I realized my mistake... oh well.
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Yes, that's the conceptual way to do it!
Just a small correction: i isn't a Gaussian prime, but a unit (a number e such that e w = 1 for some w ). Luckily, the final answer is the same in this case.
A prime factorization of 10 is 1 0 = ( 1 − i ) 2 ( 1 + 2 i ) ( 2 + i ) , and the number of factors is 4 × 3 × 2 × 2 = 4 8 , where the factor 4 accounts for the four units, ± 1 and ± i .
There is one issue to think about: We are assuming that the prime factorization is unique, up to multiplication by units (which can be proven to be the case for the Gaussian Integers). One of these days I will post an example where the factorization fails to be unique... then things get trickier.
Now you are ready for this one
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So if I understand correctly, we would not decompose 1 0 as ( 1 + i ) ( 1 − i ) ( 2 + i ) ( 2 − i ) , (even though all these terms are Gaussian primes), since we could absorb one of the units to convert one of the terms into one of the others, (e.g., − i ( 1 + i ) = 1 − i ). Since this cannot be done with the factorization that you have stated, it is the "unique" Gaussian factorization of 1 0 up to multiplication by any of the units.
(We could also write the factorization as − ( 1 + i ) 2 ( 1 + 2 i ) ( 2 + i ) , but I suppose that the convention is to write the factorization without a negative sign at the beginning.)
Thanks for posting these questions; I can't recall covering this (rather cool) topic before. :)
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@Brian Charlesworth – Thank you for taking an interest in my little problem. I learned a lot of "cool" algebra like this in high school (Gymnasium in Switzerland)... my math teacher had gotten his PhD from the great algebraist Van der Waerden, and he was fascinated by all things algebraic.
I'm not sure there is a standard answer to your questions, but Wolfram tells us that a unit up front is alright (as long as we don't count it as a prime), as in 2 = − i ( 1 + i ) 2 . Something like 2 = ( 1 + i ) ( 1 − i ) seems to be an acceptable prime factorization as well. When counting the factors, we would have to recognize that the two prime factors are "the same up to a unit."
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@Otto Bretscher – Thanks for the clarification. It sounds like you had a far more enlightening high school experience than I did. :)
Thanks. I was trying to find some theory stuff on this.
I'm impressed by your attempt to solve this by "brute force"! Thanks!
I'm not sure I fully understand your complex work, particularly your scenarios 5 and 6. Should you have c = ± 5 there? Also, I think there is some overcount; you get only 4 cases by making ∣ a ∣ = ∣ b ∣ = 1 , not the 8 you count.Lastly, you list 10 scenarios with 4 cases each, but then your total comes out to be 48?! Can you explain this a bit more?
I did a similar enumeration, just to check my conceptual work . Note that ∣ z ∣ 2 ∣ w ∣ 2 = 1 0 2 so that ∣ z ∣ 2 must divide 100. In each case we must count the ways in which ∣ z ∣ 2 can be written as the sum of the squares of two integers, ∣ z ∣ 2 = a 2 + b 2 for z = a + b i , a well-understood problem. For example, 1 0 = 3 2 + 1 2 can be written as a sum of two squares in 8 ways if you account for sign and order. ∣ z ∣ 2 = 1 z = ± 1 , z = ± i 4 cases
∣ z ∣ 2 = 2 = 1 2 + 1 2 z = ± 1 ± i 4 cases
∣ z ∣ 2 = 4 z = ± 2 , z = ± 2 i 4 cases
∣ z ∣ 2 = 5 = 2 2 + 1 2 z = ± 2 ± i , z = ± i ± 2 8 cases
∣ z ∣ 2 = 1 0 = 3 2 + 1 2 z = ± 3 ± i , z = ± i ± 3 8 cases
In the first four cases, w = z 1 will be a factor as well, with ∣ w ∣ 2 > 1 0 , so that we have to double count those cases, for a total of 4 8 factors.
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Brutal way was the only way that I knew. I still have difficulties solving part 2. I actually used a Python code to solve the problem. And then trying to explain it. A bit of foul play. I have shown all the Gaussian integer pairs above.
Oh man! how the hell could u type so much of brute force? +1
For Gaussian integer z = a + b i to be divisible by 1 0 , then there exists some Gaussian integer w = c + d i such that:
⇒ z × w = ( a + b i ) ( c + d i ) = 1 0 + 0 i ⇒ { a c − b d = 1 0 a d + b c = 0
It is obvious that a , b , c , d ∈ [ − 1 0 , 1 0 ] . Consider the cases when a d + b c = 0 .
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ a , c = b , d = 0 a d = − b c ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ a = c = 0 b = d = 0 { b , d = ± 1 , ∓ 1 0 b , d = ± 2 , ∓ 5 4 cases 4 cases { a , c = ± 1 , ± 1 0 a , c = ± 2 , ± 5 4 cases 4 cases ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ a , b = − 1 ; b , a = ± 1 a , b = + 1 ; b , a = ± 1 a , b = − 1 ; b , a = ± 2 a , b = + 1 ; b , a = ± 2 a , b = − 1 ; b , a = ± 3 a , b = + 1 ; b , a = ± 3 a , b = − 2 ; b , a = ± 4 a , b = + 2 ; b , a = ± 4 c = − 1 , d = ∓ 5 c = + 1 , d = ∓ 5 c = − 2 , d = ∓ 4 c = + 2 , d = ∓ 4 c = − 1 , d = ∓ 3 c = + 1 , d = ∓ 3 c = − 1 , d = ∓ 2 c = + 1 , d = ∓ 2 4 cases 4 cases 4 cases 4 cases 4 cases 4 cases 4 cases 4 cases
Therefore, there are altogether 4 8 Gaussian integers z that divide 1 0 (see table below).