Can Any Integer Congruent to 3 Modulo 4 Be Expressed as a Sum of Two Squares?

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Homework Statement


All work is done in the set of integers, Z
a is congruent to 3 modulo 4, Want to prove a=/= c^2 + d^2 for any c or d

Homework Equations


The Attempt at a Solution


My work:
a congruent to 3 modulo 4 => a -3 = 4k
or a = 4k +3
Then want to prove that: 4k+3 =/= c^2 + d^2 *** for any c or dI've tested out some experimental values for 4k+3 and indeed none of them have any c or d that match this equation but I'm not sure how to progress from *** onwards, It is probably a very simple matter but I can't wrap my head around this.
 
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pandaBee said:

Homework Statement


All work is done in the set of integers, Z
a is congruent to 3 modulo 4, Want to prove a=/= c^2 + d^2 for any c or d

Homework Equations





The Attempt at a Solution


My work:
a congruent to 3 modulo 4 => a -3 = 4k
or a = 4k +3
Then want to prove that: 4k+3 =/= c^2 + d^2 *** for any c or d


I've tested out some experimental values for 4k+3 and indeed none of them have any c or d that match this equation but I'm not sure how to progress from *** onwards, It is probably a very simple matter but I can't wrap my head around this.

You can express ANY integer c as 4k, 4k+1, 4k+2 or 4k+3. What are the possible expressions for c^2? In other words, what are the possible values of c^2 mod 4?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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