Gaussian Numbers (Math help needed)

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Hmmm, challenge accepted! (I suck at explaining things :P)
In Z/4Z there are only 4 elements {0,1,2,3}. (Is "elements" the proper english term?)
0^2=0
1^2=1
2^2=4=0 (mod 4)
3^2=9=1 (mod 4).
So a^2+b^2 has to be either 0+0=0 or 0+1=1 or 1+1=2, it can't be 3. Thus, if n=a^2+b^2 it can't be =3 (mod 4).
Is this making any sense? XD

Definition: The symbol Z means "The integers"
Definition: not sure if this helps but you are dealing with mod 4. the operator "mod" means you take the remainder of what it is after you divide. So 3^2 mod 4 = 9 mod 4 = 1, because 9/4 = 2 remainder of 1.

The nature of this problem deals with integers and mod 4. When the problem says "if n (operator) 3", this means that what you should assume is "suppose n mod 4 = 3".

So that's what you do. You suppose the "if" part.

Suppose n is an integer, and n mod 4 = 3.

You know that n can only be one of the following: 0, 1, 2, and 3. This is because when you take any integer and divide it by 4, the only possible remainders can be 0, 1, 2, or 3.

Then you suppose, for contradiction, that n = a^2 + b^2, where a and b are integers.

[Essentially through this contradiction you are going to prove that the sum of two squares when divided by 4 will never have a remainder of 3]

Well let's consider the squares in Z/4Z. As I said the only possible numbers you can have are 0, 1, 2, and 3.

0^2 mod 4 = 0. Okay. Next one
1^2 mod 4= 1. Cool.
2^2 mod 4 = 4 mod 4 = 0. therefore 2^2 = 0.
3^2 mod 4 = 9 mod 4 = 1. therefore 3^2 = 1.

So you have just proven that the only possible digits that can exist as a square in Z/4Z are 0 and 1. Therefore the binary sum of 0 and 1 can never result in 3. In fact, the highest possible value you can get is 1 + 1 = 2. It can never be 3.

Wow Magma Claw, I didn't think you'd get any replies here on SG but it seems there are some smarties here after all! :D

Now we won't have to post all over reddit for you.

Congratulatory bump!

I'm going to sleep now, so I guess I'll be missing out on those free keys :P

good luck with what you're working on! :D sadly I can't be of much help :(

Without reading the proof, it took me exactly 0.1 seconds to realize that 3 cannot be the sum of two squares.

Show that 6-1 is a Gaussian prime using the formula given in question 2.
Show that if m+ni divides ordinary number k, then m-ni divides k as well.

What is 6-1? What formula was given in question 2? What are m,n,i?