introduction to number theory

if \(a\) and \(n\) are divisible by a number which \(b\) is not divisible by, then there will be no solutions. Of course then we could observe that. \end{equation*}. 1 It is possible to develop a mathematical theory of partitions, prove statements about all partitions in general and then apply those observations to our case here. }\) Of course \(ad\) is divisible by \(d\text{,}\) as is \(bd\text{. We could do long division, but there is another way. }\) Instead, we write \(8 \equiv 23\text{. \( \def\circleAlabel{(-1.5,.6) node[above]{$A$}}\) Published October 2005,February 2011. 5.2: Introduction to Number Theory Divisibility. It works if we are thinking division by 5, so we need to denote that somehow. \( \def\AAnd{\d\bigwedge\mkern-18mu\bigwedge}\) }\) Here we are looking for all the numbers divisible by \(5\) since \(a = 5q+0\text{. Write your solution as an equation, such as, \begin{equation*} x = n + kb \end{equation*}, Plug this into the original Diophantine equation, and solve for \(y\text{.}\). Pick the smaller of \(a\) and \(b\) (here, assume it is \(b\)), and convert to a congruence modulo \(b\text{:}\) \begin{equation*} ax + by \equiv c \pmod{b}. }\), \(\renewcommand{\bar}{\overline}\) Think of congruence as being “basically equal.” If we have two numbers which are basically equal, and we add basically the same thing to both sides, the result will be basically equal. \( \def\circleBlabel{(1.5,.6) node[above]{$B$}}\) An equation in two or more variables is called a Diophantine equation if only integers solutions are of interest. \( \def\sat{\mbox{Sat}}\) To be safe, let's divide as much of \(n\) as we can. \( \def\threesetbox{(-2.5,-2.4) rectangle (2.5,1.4)}\) In particular, if we divide both sides by 17, we must get the same remainder. While 20 is a multiple of 4, it is false that \(4\) is a multiple of 20. We know that \(a \equiv b \pmod{n}\) if and only if \(n \mid a-b\text{,}\) if and only if \(a-b = kn\) for some integer \(k\text{. \( \def\inv{^{-1}}\) }\) So \(kn\) must also be divisible by \(d\text{. The most important fact about partitions, is that it is possible to define an equivalence relation from a partition: this is a relationship between pairs of numbers which acts in all the important ways like the “equals” relationship. Try 83: Is this the best we can do? It matters what the divisor is: \(8\) and \(23\) are the same up to division by \(5\text{,}\) but not up to division by \(7\text{,}\) since \(8\) has remainder of 1 when divided by 7 while 23 has a remainder of 2. This would be going too far, so we will refuse this option. }\) In other words, if \(b \mid a\text{,}\) then \(a = bk\) for some integer \(k\) (this is saying \(a\) is some multiple of \(b\)). In fact, we now know something more: any number is congruent to the sum of its digits, modulo 9. }\) If it spits out anything other than an integer, you know \(m \nmid n\text{. Then you can start reading Kindle books on your smartphone, tablet, or computer - no Kindle device required. Thus we can simplify further: \begin{equation*} 6^{41} = 6\cdot (6^2)^{20} \equiv 6 \cdot 1^{20} \pmod 7. First, check if perhaps there are no solutions because a divisor of \(51\) and \(87\) is not a divisor of \(123\text{. It is easy to add and multiply natural numbers. We want to say that \(8\) and 23 are basically the same, even though they are not equal. There was a problem loading your book clubs. }\) Again we get, \begin{equation*} 8 \equiv 13 \pmod 5. One of the important consequences of these facts about congruences, is that we can basically replace any number in a congruence with any other number it is congruent to. How can you make $6.37 using just 5-cent and 8-cent stamps? \( \def\shadowprops, \( \newcommand{\hexbox}[3]{ \end{equation*} So the solutions are those values which are congruent to 8, or equivalently 3, modulo 5. 54, No. If we subtract, we get \(136299 - 136286 = 13\text{. \( \def\circleA{(-.5,0) circle (1)}\) \end{aligned} \end{equation*}. We get the infinite set. Having the most stamps means we have as many 5-cent stamps as possible, and to get the smallest number of stamps would require have the least number of 5-cent stamps. \( \def\O{\mathbb O}\) Note that in the example above, every integer is in exactly one remainder class. To minimize the number of 5-cent stamps, we want to pick \(k\) so that \(121-8k\) is as small as possible (but still positive). \end{equation*} This will reduce to a congruence with one variable, \(x\text{:}\) \begin{equation*} ax \equiv c \pmod{b}. Well, certainly 1, does, as does 6, and 11. True. This is just a way of saying it using multiplication. … The authors succeed in presenting the topics of number theory in a very easy and natural way, and the presence of interesting anecdotes, applications, and recent problems alongside the obvious mathematical rigor makes the book even more appealing. This means that \(x = 9\) and \(x = 14\) and \(x = 19\) and so on will each also be a solution because as we saw above, replacing any number in a congruence with a congruent number does not change the truth of the congruence.