This is actually two questions:
1.Do there exist numbers of the form $n^2+1$ with arbitrarily many unique prime factors?
2.Do there exist numbers of the form $n^2+1$ with arbitrarily many prime factors?
I think an approach using gaussian primes might help,
$$n^2+1=(n+i)(n-i)$$
let $n=c^2$
$$(c^2 + i)(c^2-i)...$$
but from here I haven't managed to make anything productive out of this approach.
Note: the primes in the question are normal natural number primes.
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razivo
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What do you mean by unique? – Nov 27 '20 at 07:13
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different prime factors, i.e. $12$ has two unique prime factors $2,3$, and three prime factors $12=2\cdot 2\cdot 3$ – razivo Nov 27 '20 at 07:15
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1Oh ok, usually when we talk about orime factors, if $n=p_1^{a_1}\cdotp_2^{a_2}....\cdotp_m^{a_m}$, $n$ has $m$ prime factors. – Nov 27 '20 at 07:16
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Yeah, I probably should have mentioned it. – razivo Nov 27 '20 at 07:19
2 Answers
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Yes and yes. In fact, given any finite set of primes $p_i \equiv 1 \pmod 4$, there exists an integer $n$ such that $n^2 + 1$ is divisible by all of the $p_i$. Proof: each of the congruences $n_i^2 \equiv -1 \pmod{p_i}$ individually has two solutions modulo $p_i$; choose one of the two for each $i$, and use the Chinese Remainder Theorem to find an $n$ congruent to each $n_i$ modulo $p_i$.
Ravi Fernando
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can you elaborate? why does $n_i^2\equiv-1 \pmod {p_i}$ have solutions? is it because $p_i\equiv1 \pmod 4$ – razivo Nov 27 '20 at 07:10
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1@razivo .If $G$ is a finite commutative group with identity $1$ then $\prod_{x\in G}x=\prod_{x\in S}x$ where $S={x\in G: x^2=1}$ because the other members of $G$ cancel out in pairs. In particular, if $p$ is a prime and $G={1,...,p-1}$ with multiplication mod $p$ then $S={1,p-1}$ so $(p-1)!\equiv -1 \mod p.$ (Wilson's Theorem.) So if also $4|(p-1)$ then with $p=4c+1$ we have $-1\equiv (p-1)! \equiv (2c)!\prod_{j=1}^{2c}(p-j)\equiv (2c)!^2 \mod p.$ – DanielWainfleet Nov 27 '20 at 11:46
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More discussion of this theorem, including Daniel's proof and some others: https://math.stackexchange.com/questions/122048/1-is-a-quadratic-residue-modulo-p-if-and-only-if-p-equiv-1-pmod4 and https://math.stackexchange.com/questions/2731641/if-p-is-prime-and-p-equiv-1-mod-4-then-the-congruence-x2-equiv. – Ravi Fernando Nov 27 '20 at 18:28
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I will elaborate what Ravi said for you.
$n_i^2\equiv -1\pmod{p_i}$ has a solution if $-1$ is a quadratic residue. A known formula is $$\bigg(\frac{-1}{p}\bigg)=(-1)^\frac{p-1}{2}$$ thus, if and only if $p\equiv 1\pmod{4}$ $-1$ is a quadratic residue.
(search up quadratic residues and laws for more information).
Now simply consider $n$ according to the Chinese Remainder Theorem.