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# Tag Archives: Number Theory

## 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

## Irreducible fraction

Find all positive integers such that it holds that

(1)

where stands for the period.

**Solution**

Well,

## Divisibility

Prove that the product of consecutive positive integers divides .

**Solution**

and the result follows.

*Any questions?*

## gcd and subgroup

Let be a non trivial subgroup and . Prove that

**Solution**

We have that for any integers and . But by Bezout’s lemma we can find such that

and hence .