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# Tag Archives: Number Theory

## On the factorial

Let denote the Möbius function and denote the floor function. Prove that:

Solution

The RHS equals

since for .

## Proof of “Fermat’s last theorem”

Let and . Prove that the equation

has no solution.

Solution

Without loss of generality , assume that . If held , then it would be thus . It follows from Bernoulli’s inequality that,

which is an obscurity. The result follows.

## Eulerian equality

We know that there are infinite Pythagorian triplets, that is numbers such that

(1)

Let us investigate if there exist triplets such that

(2)

where denotes the Euler’s totient function.

Solution

Indeed, there are infinite triplets such that is satisfied. For example noticing that

we deduce that for each natural such that we have

## Square of an integer

Let be positive integers such that divides . Show that

is the square of an integer.

Solution

This is a very well known problem. It first appeared in IMO held in Canberra, Australia. It even has its own Wikipedia page.

History Background

One of the organisers had sent this exercise to all professors of Number Theory to check if the exercise is original and try to solve it within hours. No one was able to do so. However, one Bulgarian student managed to solve the problem in less than hours ( his solution was fallen from the sky )  and for that he was awarded a special reward beyond the medal.

Solution

Choose integers such that Now, for fixed , out of all pairs choose the one with the lowest value of . Label . Thus, is a quadratic in . Should there be another root, , the root would satisfy: Thus, isn’t a positive integer (if it were, it would contradict the minimality condition). But , so is an integer; hence, . In addition, so that . We conclude that so that .

This construction works whenever there exists a solution for a fixed , hence is always a perfect square.

## On permutation

Let be vectors of – dimensional Euclidean space such that . Prove that there exists a permutation of the integers such that

for each .

Solution

We define inductively. Set .Assume is defined for and also

(1)

Note that is true for . We choose in a way that is fulfilled for instead of . Set and . Assume that for all .  Then and in view of ones gets which is impossible. Hence , there is such that

(2)

Put . Then using and we have

which verifies for . Thus we define for every . Finally from we get

### Who is Tolaso?

Find out more at his Encyclopedia Page.