I try to add something more into the set of notes. All prime numbers, except 2, have to be in the form of 4k+1 or 4k+3 (because all the other primes are odd). Two natural follow-up questions are, "Is there infinite number of primes of the form 4k+3?" and "Is there infinite number of primes of the form 4k+1?" The proof can be included in the set of notes as an appendix for more gifted students.
It turns out that the first question is easy to answer, by mimicking the technique from Euclid. If you have never heard this before, try to understand the proof by Euclid from the link above, and adopt the main idea to the "4k+3" question.
The second question, however, is more difficult. I am not sure if there is any simpler elementary solution, but I found a rather fascinating proof on web, using the Fermat's Little Theorem. Let me restate the proof here.
Given any integer N, consider
Since 1, 2, ... , N all do not divide M, p is larger than N. Since p divides
Taking
On the other hand, by Fermat's Little Theorem, for a = 1, 2, ... , N, we have
Multiplying the above formula with a = 1, 2, ... , N, we get
Comparing (1) and (2), we get
which implies
Is that too difficult for F.3 and F.4 students? I am not sure, but as it is put in the appendix, it is not demanding even I put the proof of Fermat's Last Theorem there. >.<""
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Is the assumption 'p is the SMALLEST prime factor' used in the proof?
Yes, you are correct! Actually p can be any prime factor of (N!)^2+1. That sounds a more amazing result.
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