Exercice

Find all pairs

of positive integers such that

and

are coprime and

divides

,

, if there exist

such that

and then

, but

but

and

are coprime

,

. contradiction .

now we have

and we have and

, and

For n real numbers

let

are natural numbers because

and

are natural .

denote the difference between the greatest and smallest of them and

Prove that

and find when each equality holds.

Let

be a sequence so that

every pair of sequence

. Considerer sum

are counted exactly once

is counted

and arbitrary

times. This implies that we can permute the indices of

as we wish.

WLOG we can thus assume

and write

.

Now define a sequence In the sum

. Because

.

appears

times,

appears

The sum becomes

times and so on, in general

appears

times.

.

Now let's see what happens if we modify

by some length m.

.

On the other hand

.

Examining the difference

Assuming when

.

we have

when

, strengthening the inequality and similarly weakening the inequality

.

Obviously when we weaken the inequality to its limit

for every j and

To construct every possible situation we can start from

and start adding some m to them.

To keep the equality we must have That is, only

.

Resulting

.

, if it exists, may be greater than zero for the equality.

Now considering the other inequality,

:

We have

.

Like above, we have

\ge 0 when

Again, at its limit

.

Here for the equality we want That is, only

and

, strengthening the inequality and weakening when

, which implies

may be greater than zero for the equality.

or

.

.