# A Course on Number Theory [Lecture notes] by Peter J. Cameron

By Peter J. Cameron

**Read Online or Download A Course on Number Theory [Lecture notes] PDF**

**Best number theory books**

This e-book constitutes the 5th and ultimate quantity to set up the consequences claimed via the good Indian mathematician Srinivasa Ramanujan in his "Notebooks" first released in 1957. even though all the 5 volumes includes many deep effects, might be the common intensity during this quantity is bigger than within the first 4.

**Problem-Solving and Selected Topics in Number Theory: In the Spirit of the Mathematical Olympiads**

This ebook is designed to introduce probably the most very important theorems and effects from quantity thought whereas trying out the reader’s knowing via rigorously chosen Olympiad-caliber difficulties. those difficulties and their recommendations give you the reader with a chance to sharpen their talents and to use the idea.

**Primality testing for beginners**

How will you inform no matter if a bunch is fundamental? What if the quantity has 1000s or millions of digits? this question could appear summary or inappropriate, yet actually, primality exams are played at any time when we make a safe on-line transaction. In 2002, Agrawal, Kayal, and Saxena responded a long-standing open query during this context via providing a deterministic attempt (the AKS set of rules) with polynomial operating time that tests no matter if a host is fundamental or now not.

**Additional resources for A Course on Number Theory [Lecture notes]**

**Example text**

An ] You tell them the numerator and denominator of this fraction. There are a couple of traps, one theoretical, one practical. First of all, as we have seen, the greatest common divisor of the numerator and denominator of a continued fraction is 1. So if the original numbers r and s have greatest common divisor d > 1, then you will find r/d and s/d instead of r and s. There is nothing that can be done about this; you have to bluff your way out of it as well as you can. The practical problem is caused by rounding errors.

An−2 ]. Proof (a) holds because in Euler’s formula the reversed sequence obviously gives exactly the same result. Then (b) is straightforward: [a0 , . . , an ] = [an , . . , a0 ] = an [an−1 , . . , a0 ] + [an−2 , . . , a0 ] = [a0 , . . , an−1 ]an + [a0 , . . , an−2 ]. 3 = = = = = 3 [3]1 + 1 = 4 [3, 1]4 + [3] = 19 [3, 1, 4]1 + [3, 1] = 23 [3, 1, 4, 1]6 + [3, 1, 4] = 157. The convergents of a finite continued fraction Let a0 , a1 , . . , an be positive integers. We define the convergents of the continued fraction [a0 ; a1 , .

We begin with a lemma about purely periodic continued fractions. 2 If y = [a0 ; a1 , . . , an−1 , an ], then −1/y = [an ; an−1 , . . , a1 , a0 ]. Proof y = [a0 ; a1 , . . , an , y] = ypn + pn−1 , yqn + qn−1 so qn y2 + (qn−1 − pn )y + pn−1 = 0. Let z = [an ; an−1 , . . , a0 ]. Then z[an , . . , a0 ] + [an , . . , a1 ] z[an−1 , . . , a0 ] + [an−1 , . . , a1 ] zpn + qn = , zpn−1 + qn−1 z = [an ; an−1 , . . , a0 , z] = where we use the fact that [a0 , . . , an ] = [an , . . , a0 ]. So pn−1 z2 + (qn−1 − pn )z − qn = 0.