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
, and p be the smallest prime factor of M.Since 1, 2, ... , N all do not divide M, p is larger than N. Since p divides
, we have
Taking
-th power on both sides, we have
....(1)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
....(2)Comparing (1) and (2), we get
which implies
is even and hence p is of the form 4k+1. As N is chosen arbitrarily, we reach the conclusion: there are infinitely many primes of the form 4k+1.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. >.<""
= 1？」這問題。我們只需簡單的極限概念（甚至幾何級數和的概念），便知道這答案是肯定的，
，我們表示 
，其中 
是整數部分，而 
是第 
、
、
、……。例如，我們指 
時，其實指 
是 
、
、
、
、……這數列的極限。
。這就好像兩個分數式 
的值相同一樣。既然如此，為甚麼 
、
、
、……的極限是 
愈來愈接近 
為例，因為 
。