## 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.

## Nested radical inequality

Let . Prove that

Solution

The LHS is equal to which by AM – GM is less or equal to

where . Since it follows from Bernoulli inequality that .

## No invertible matrices

Show that there do not exist invertible matrices such that and .

Solution

Suppose, on the contrary, that such matrices do exist. Then

and also

Using the fact that we deduce that

The last means that which is impossible because both and are invertible ( and so must be the product ). Hence, the conclusion follows.

## Matrix determinant inequality

Suppose that all eigenvalues of are positive real numbers. Show that

Solution

Let the eigenvalues of be , . Consider the Jordan normal form of ;  this Jordan form is an upper triangular matrix that has the eigenvalues of in the main diagonal. Let this matrix be called .  Furthermore , as a matrix and its Jordan normal form are similar. As is upper triangular, its inverse is given by an upper triangular matrix whose diagonal entries are the inverses of the diagonal entries of . That is,

is an upper-triangular matrix, and its determinant can be computed simply as the product of the elements of the diagonal. Hence,

due to the inequality for all .
Note: The main point here is that if is an eigenvalue of then is an eigenvalue of . Taking into account that

yields a shorter solution to the problem. No need for upper triangular matrices.

## Root inequality

Let be positive real numbers such that . Prove that

Solution

Well if we apply AM-GM to we obtain

(1)

and similarly if we apply AM – GM to we obtain

(2)

We have successively,