Problem 556

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v6ph1
Posts: 134
Joined: Mon Aug 25, 2014 7:14 pm

Problem 556

Post by v6ph1 »

Problem 556 (View Problem)
The Factorization of 1+3i seams to be wrong:
It should be (1+i)*(2+i)
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mpiotte
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Re: Problem 556

Post by mpiotte »

v6ph1 wrote:Problem 556 (View Problem)
The Factorization of 1+3i seams to be wrong:
It should be (1+i)*(2+i)
Thanks, corrected.
Swistakk
Posts: 66
Joined: Thu Feb 08, 2024 9:16 pm

Re: Problem 556

Post by Swistakk »

mpiotte wrote: Sun Apr 17, 2016 3:28 pm
v6ph1 wrote:Problem 556 (View Problem)
The Factorization of 1+3i seams to be wrong:
It should be (1+i)*(2+i)
Thanks, corrected.
Well, not really...
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echip
Posts: 39
Joined: Sat Feb 15, 2014 6:51 pm

Re: Problem 556

Post by echip »

mpiotte wrote: Sun Apr 17, 2016 3:28 pm
v6ph1 wrote:Problem 556 (View Problem)
The Factorization of 1+3i seams to be wrong:
It should be (1+i)*(2+i)
Thanks, corrected.
You say it's corrected, But I see (1+i)*(1+2*i)
skoczian
Posts: 28
Joined: Sat May 25, 2013 4:43 pm

Re: Problem 556

Post by skoczian »

mpiotte wrote: Sun Apr 17, 2016 3:28 pm
v6ph1 wrote:Problem 556 (View Problem)
The Factorization of 1+3i seams to be wrong:
It should be (1+i)*(2+i)
Thanks, corrected.
This factorization occurs twice in the problem text. Only once it's corrected:
A Gaussian integer can be uniquely factored as the product of a unit and proper Gaussian primes.
For example 2 = -i(1 + i)2 and 1 + 3i = (1 + i)(2 + i).
Here it's correct.
And in the next paragraph:
For example f(10) = 7 because 1, 1 + i, 1 + 2i, 1 + 3i = (1 + i)(1 + 2i), ...
Still wrong.
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RobertStanforth
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Re: Problem 556

Post by RobertStanforth »

Thank you everyone for flagging this. The second occurrence has now been corrected to read (1+3i)=(1+i)(2+i).
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