Prove or disprove $$\dfrac{n^2 + 1}{4k + 3} \notin \mathbb{Z}, n, k \in \mathbb{N}^{+}$$
I know if $n^2 + 1$ is prime if and only if $n^2 + 1 \equiv 1 \pmod 4$.
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By $N^+$, do you mean the set of positive integers $\Bbb{Z^+}$ ? – Prasun Biswas Apr 10 '15 at 12:06
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Yes,That's same meaning – Apr 10 '15 at 12:06
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3This question seems to answer your question. – mathlove Apr 10 '15 at 12:12
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Suppose $\,4k\!+\!3\mid n^2\!+1.\,$ $\,4k\!+\!3\,$ has a prime factor $\,p\equiv 3\pmod 4\,$ [else all are $\equiv 1\,$ so their product $\equiv 1,\,$ not $\, 3\,$]. Write $\, p = 4j\!+\!3,\,$ therefore $\, \color{#0a0}{p\!-\!1 = 2(2j\!+\!1)},\,$ so by little Fermat
$p\mid 4k\!+\!3\mid \color{#90f}{n^2\!+\!1}\,\Rightarrow\,{\rm mod}\ p\!:\,\ \color{#90f}{n^2\equiv -1}\ \Rightarrow\ \color{#c00}1\equiv n^{\color{#0a0}{p-1} }\equiv (\color{#90f}{n^2})^{\color{#0a0}{2j+1}}\equiv (\color{#90f}{-1})^{2j+1}\equiv \color{#c00}{-1}$
therefore $\ p\mid 2=\color{#c00}{1-(-1)},\ $ contra $\,p = 4j+3\,$ is odd.
Bill Dubuque
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