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Gaussian integers and factorisation over Gaussian primes

So I came across a proof for a^2+b^2+c^2=d^2 but it requires knowledge of gaussian integers and factorisation over gaussian primes. Could anyone explain to me what they are or link any resources(I couln't find much on this topic without getting too complicated for me). Thnx
Reply 1
Original post by 11234
So I came across a proof for a^2+b^2+c^2=d^2 but it requires knowledge of gaussian integers and factorisation over gaussian primes. Could anyone explain to me what they are or link any resources(I couln't find much on this topic without getting too complicated for me). Thnx


Don't have a link to anything but can probably quickly sum up the main points you'll need to know.

Gaussian integers are complex numbers of the form a+bi where a and b are integers and i^2 =-1.

The sum, difference and product of two GIs is a GI.

The only GIs with a multiplicative inverse (the units) are 1, -1, i and -i.

The GIs form a unique factorization domain - this means there is a "fundamental theorem of arithmetic" for them and every Gaussian integer can be factorized uniquely into prime GIs.

Note an everyday prime need not be a GI prime - e.g. 2 = (1+i)(1-i).

Note that also 2 = (i-1)(-i-1) but this is still the same factorization essentially - we've just moved around some units. (Like 6 = 2 x 3 = (-2) x (-3).)

Hope this helps.
Reply 2
Original post by RichE
Don't have a link to anything but can probably quickly sum up the main points you'll need to know.

Gaussian integers are complex numbers of the form a+bi where a and b are integers and i^2 =-1.

The sum, difference and product of two GIs is a GI.

The only GIs with a multiplicative inverse (the units) are 1, -1, i and -i.

The GIs form a unique factorization domain - this means there is a "fundamental theorem of arithmetic" for them and every Gaussian integer can be factorized uniquely into prime GIs.

Note an everyday prime need not be a GI prime - e.g. 2 = (1+i)(1-i).

Note that also 2 = (i-1)(-i-1) but this is still the same factorization essentially - we've just moved around some units. (Like 6 = 2 x 3 = (-2) x (-3).)

Hope this helps.

Thanks, how could I use this to prove fermats theorem. Is there any way to get from gaussian integers back to normal integers
Reply 3
Original post by 11234
Thanks, how could I use this to prove fermats theorem. Is there any way to get from gaussian integers back to normal integers


The following is a broken down proof of Fermat's Theorem which may help.

Let p=4n+1 be a prime in Z.

Prove that (p-1)! -1 mod p.

Prove that ((2n)!)² (4n)! mod p.

Hence show that

((2n)!-i)((2n)!+i) is a multiple of p in Z

and that p is not prime in Z.

Deduce that p is the sum of two squares in Z.

Can a prime of the form 4n+3 be expressed as the sum of two squares?

[Here Z denotes the integers and Z the Gaussian integers.]

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