<s>[INST] $n^2 - n + 11$ is the product of four prime numbers, some of which may be the same. What is the minimum value of $n$? Let $n$ be a positive integer such that $n^2 - n + 11$ is the product of four prime numbers, some of which may be the same. What is the minimum value of $n$?
The problem is taken from here. I know the answer is $132$, but how can I prove it? That would mean the four numbers are $11$, $11$, $11$, and $13$.
Can you give me a hint, please? Thanks!
 [/INST] First, find the smallest prime that divides $n^2-n+11$. It is not 2:
$$0^2-0+11 \equiv 1 \pmod{2}$$
$$1^2-1+11 \equiv 1 \pmod{2}$$
It is not 3:
$$0^2-0+11\equiv 2\pmod{3}$$
$$1^2-1+11\equiv 2\pmod{3}$$
$$2^2-2+11\equiv 1\pmod{3}$$
Not 5:
$$0^2-0+11\equiv 1\pmod{5}$$
$$1^2-1+11\equiv 1\pmod{5}$$
$$2^2-2+11\equiv 3\pmod{5}$$
$$3^2-3+11\equiv 2\pmod{5}$$
$$4^2-4+11\equiv 3\pmod{5}$$
Not 7:
$$0^2-0+11\equiv 4\pmod{7}$$
$$1^2-1+11\equiv 4\pmod{7}$$
$$2^2-2+11\equiv 6\pmod{7}$$
$$3^2-3+11\equiv 3\pmod{7}$$
$$4^2-4+11\equiv 2\pmod{7}$$
$$5^2-5+11\equiv 3\pmod{7}$$
$$6^2-6+11\equiv 6\pmod{7}$$
This means the smallest prime that divides $n^2-n+11$ is 11. So, you use trial and error:
$$n^2-n+11=11^4$$
has no solutions over the integers
$$n^2-n+11=11^3\cdot 13$$
Here you get 132.
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