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Let denote the Möbius function and denote the floor function. Prove that:
The RHS equals
since for .
Let and . Prove that the equation
has no 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.
We know that there are infinite Pythagorian triplets, that is numbers such that
Let us investigate if there exist triplets such that
where denotes the Euler’s totient function.
Indeed, there are infinite triplets such that is satisfied. For example noticing that
we deduce that for each natural such that we have
Let be positive integers such that divides . Show that
is the square of an integer.
This is a very well known problem. It first appeared in IMO held in Canberra, Australia. It even has its own Wikipedia page.
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.
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.
Let be vectors of – dimensional Euclidean space such that . Prove that there exists a permutation of the integers such that
for each .
We define inductively. Set .Assume is defined for and also
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
Put . Then using and we have
which verifies for . Thus we define for every . Finally from we get