Math Olympiad Preparation II

In this post I am going to talk mainly about how to prepare for Math Olympiad. I am going to discuss two different approaches.  At the bottom of this post are two videos I created with more information on me, Math Olympiad prep, and my online tutoring services. The first method is to start off with solving Olympiad-type questions. You can, for example, find past IMO exams here and see how well you can do. There are many other websites and books containing Olympiad-type problems and solutions. In this method I'd suggest you start with old exams. They tend to be much easier. In the process, you will find out which subjects you need to(more...)

Solution of FLT for $n=4$.

I discussed FLT in an earlier post and stated a problem that if solved FLT for will be deduced from there. Below is a solution to that problem. Assume is a “non-trivial” triple of integers that satisfies (1) . By “non-trivial” we mean “". Obviously we can assume that and are positive. If we show for any such triple, there is another triple satisfying the same equation with a smaller , then using the method of “infinite descent” we get a contradiction, and therefore we can deduce that there is no “non-trivial” solutions for . Now one can show that and are relatively prime. Otherwise you can find a “smaller” solution for the equation (1).(more...)

Mathematics Olympiad Preparation

International Mathematics Olympiad is a highly respected competition which is held yearly. It is a place where many mathematicians are hoping to motivate the best students to continue their studies in Math. A lot of participants end up becoming the best mathematicians later in their career. My Ph.D. adviser Professor Gregory Margulis who won the Fields Medal in 1978- among many other prestigious Math prizes- won a silver medal in the 4th IMO. Two other examples of these mathematicians are 2006 and 2010 winners of Fields Medal, Terence Tao and Elon Lindenstrauss. Different countries have different policies toward IMO. Some countries like China and my home country Iran train their teams for a long time, up(more...)

Fermat's Last Theorem

One of the most important achievements of the 20th century in Math is a proof of the Fermat's Last Theorem. It states: Fermat's Last Theorem. Let and be 4 integers such that and . Then . In our previous post we discussed all solutions of this equation for as Pythagorean Triples. Proof of FLT involves advanced Mathematics. Here I would like to bring up a especial case of this theorem for . More generally one can prove the following. Problem. If for 3 integers and , then . One can prove the above statement using infinite descent and combining it with Pythagorean Triples. Think about it if you get a chance. A proof will be posted later.

Pythagorean Triples

Any three integers and that can be lengths of sides of a right triangle is a called Pythagorean triple. But we can also think of negative numbers as Pythagorean triples when they satisfy the equation . Here I would like to show you a simple method to find all Pythagorean triples. Asssume is the greatest common divisor of and , i.e. . If we divide and with obviously we get another Pythagorean triples. Therefore to find all Pythagorean triples it is enough to find all triples with g.c.d.=1.  Therefore we may assume and have no common factors.  One can check that one of or has the same parity as .  (why?) Assume and have the(more...)

Infinite Descent

Infinite descent is a common mathematical method that we use in solving problems and proofs. It is based on a very simple and intuitive fact: every non-empty set of natural numbers has a smallest number. This means if you show a set of natural numbers has no smallest number that set should be empty. This method is commonly used to attack some problems in number theory. Especially for finding solutions to some specific Diophantine equations. I understand that wikipedia is not the most reliable source of information out there, but a lot of their articles are very useful, so I suggest you take a look at this wiki article to find out more about Infinite(more...)

Solution of the previous problem

Here is the solution to the problem that I posted earlier. By Sylow Theorem we know that the number of -Sylow subgroups should divide . On the other hand this number is congruent to modulo . So, we have either or , Sylow subgroups. If there is no normal subgroup of order we should have  Sylow subgroups. Let be the set of all -Sylow subrgoups of . One can check the number of elements in  is . Notice that all 's are conjugate. Therefore, we have , where is the normilizer of in . Hence, , which means for any . Take an element outside of all 's and take an element . We have . One may check(more...)

Some notations and basics

Before posting my solution to the first problem I thought I'd post some basic facts and notations that I am going to use. If you are not familiar with these basic facts it is good to check them. Everybody should check these at least once in their lifetime. Let be a group and its subgroup. denotes the set of all elements such that . One can check that is a subgroup of containing .  is called a conjugate of . You can check that the number of conjugates of is .  is the set of all elements in that commute with . One can check this forms a subgroup of . Furthermore, the number of conjugates of  equals(more...)

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