Show that if n = 3 (mod 4), then n cannot be the sum of the squares of two integers
Answers
Step-by-step explanation:
Well, assume a^2 + b^2 = 3 (mod 4).
Then a^2 + b^2 is odd, which means exactly one of a, b is odd. We can call a odd, without loss of generality. b is therefore even.
Clearly, b^2 = 0 (mod 4). The square of any even number is 0 mod 4.
Therefore, a^2 must be 3 mod 4. Ergo
a^2 = 3 (mod 4)
<=>
a^2 = 4n + 3
Any odd number can be written as either 4m+1 or 4m-1. So
(4m+1)^2 = 16m^2 + 8m + 1 = 4n + 3
<=>
16m^2 + 8m = 4n + 2
<=>
8m^2 + 4m = 2n + 1
<=>
4(2m^2 + m) = 2n + 1
A clear contradiction. Checking the other case,
(4m-1)^2 = 16m^2 - 8m + 1 = 4n + 3
<=>
16m^2 - 8m = 4n + 2
<=>
8m^2 - 4m = 2n + 1
<=>
4(2m^2 - m) = 2n + 1
Also a clear contradiction. Therefore, it cannot possible be true that
a^2 = 4n + 3
So it follows that it is not the case that
a^2 = 3 (mod 4)
From which one can deduce that, since b^2 = 0 (mod 4),
a^2 + b^2 = 3 (mod 4)
Cannot happen.
It may not be the prettiest proof, but I'm sure the other regulars will swoop in with a two-liner and make me look like a muppet. ;D
Answer:
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Step-by-step explanation:
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