<s>[INST] Argument for why $a^2 + 1$ is never divisible by a $3 \mod 4$ integer How do you show that $$a^2 + 1$$ is never divisible by a $3 \mod 4$ integer (which is equivalent to showing that it has no $3 \mod 4$ prime factor) for any non-negative integer $a$ by analysing the arithmetic series representation of $a^2$, $1 + 3 + 5 + ... + (2a-1)$?
 [/INST] Not using the series: 


*

*Case 1: $a$ is even, say, $a=2k$. Then $a^2+1 = 4k^2 + 1$.

*Case 2: $a$ is odd, say, $a=2k+1$. Then $a^2+1 = 4k^2+4k+2$.


Using the series $a^2=1+3+5+...+(2a-5)+(2a-3)+(2a-1)$. 


*

*Note that consecutive terms add to multiples of $4$. 

*Therefore, $a^2$ is either $(1+3)+(5+7)+...+((2a-3)+(2a-1))$, which is a sum of multiples of 4, or $1 + (3+5)+(7+9)+...+((2a-3)+(2a-1))$, which is one more than a multiple of $4$.

*in any case, the result follows for $a^2+1$


Explicitly in $\mathbb{Z}_4$:


*

*$a$ is either 0, 1, 2 or 3. Therefore $a^2$ is either 0 or 1, so $a^2+1$ is 1 or 2.

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