## 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

## Does there exist an expression?

For which positive integers does there exist expression

where each is a disk of radius such that each point belongs to either the boundary of some or to precisely interiors of the sets , , ?

(Euler Competition , 2017)

Solution [official]

For none . Assume , on the contrary , that such expression exists for some value of . Let us first note that the set of disks is countable hence the Lebesgue measure of the union of all the circles is .

Further, consider any disk and let us show that the number of the other disks it intersects is finite. Indeed, each disk that intersects would have to be a subset of the disk or radius co-centric with . All such disks together would cover an area of at most so that each point is covered by at most disks. Hence the total area of disks that intersect is at most and therefore there are not more than of them.

Now, let us introduce the function whose value is the number of disks that belongs to. Note that

• for almost all
• is constant on every connected component of .

What remains to be shown is that this is impossible. Since the disk intersects only with finitely many other disks , let us consider the circular arc between two such intersection points such that the arc in the figure above. But then obviously the value for points on one side of the arc and for points on the other side of the arc differ by . For the collections of disks that cover the two regions is different – is only on one side of the circular arc.

## A divergent series

Let be a positive and strictly decreasing sequence such that . Prove that the series

diverges.

Solution

We shall begin with a lemma.

Lemma: Let . It holds that

Proof: We are using induction on . For it is trivial. Suppose that it holds for . Then

Thus it holds for and the lemma is proved. Since and I can find such that for all . But then

where in the first inequality the lemma was used.

The exercise can also be found at mathematica.gr . It can also be found in Problems in Mathematical Analysis v1 W.J.Kaczor M.T.Nowak as exercise 3.2.43 page 80 .

## Seemous 2017/4

(a) Let . Evaluate the integral:

(b) Let and let be a sequence defined as

Find the limit of .

Solution

(a) We have successively:

where denotes the lower incomplete Gamma function.

(b) We are invoking Tonelli’s theorem. Hence the limit we seek is equal to