The only one is 26

Introduction

It already does enough time we comment on a curious property of the number 26. Specifically it is this one:

The number 26 is the only natural number that is placed between a square () and a bucket ().

Apparently it was Fermat who demonstrated the above mentioned result, but in the post where we realized of this characteristic of 26 no test of this fact was happening. It was Juanbuffer who was contributing in a comment a pdf with a demonstration of the same one (that if I do not remember badly was not in Spanish). Unfortunately it seems that it is already not possible to gain access to the above mentioned document (at least I cannot). For this motive I started looking … and I have found it. My admired Carlos Ivorra is who has provided the above mentioned test to me. Very well, in fact I do not know if it is his, but it appears in one of the books in format pdf that it has available in his web: Theory of Numbers.

In this article you are going to be able to see this demonstration.

The unicidad of 26

In fact the demonstration that I am going to present to you of the fact of which the only natural number is 26 with the mentioned property previously is relatively elementary. The interesting of the test is that it leaves of the set of the natural numbers to demonstrate a characteristic in. The fact to rest on a set major that to demonstrate something in him is a quite useful, made argument of which many mathematicians took advantage when they became convinced of the potency of the above mentioned argument.

Centrémonos in the topic. We are going to do the demonstration in (the entire numbers). Then the statement of the result to demonstrate the following one:

Theorem:

The only entire solutions of the equation

they are.

Demonstration:

A simple glance to the equation says to us that it cannot be an even number. If out we would have that also it would be a pair. The contradiction would be in the fact that the right part of the equality would be divisible between 8, but the left part would not be not even divisible between 4. Therefore it has to be an odd number.

We leave now of for penetrating into the ring. We think that the previous equation in this ring his expression can happen factorizada of the following way:

We consider in this ring the following norm:

It is simple to verify that the above mentioned norm is multiplicativa, this is, that it is positive for any element different from zero of, that zero is zero for the element and that the norm of a product of two elements of es the product of the norms of saying elements.

Let's suppose now that they fulfill the initial equation and let's take the elements and of. Any element that is their common divisor two must divide also to his sum, and to his difference. Taking norms in this situation we would have the following thing:

Therefore. The only pairs of values that fulfill this are them following:

With the first two possibilities we obtain the elements and-1$ of, that are units of this ring. In other cases we obtain the elements and, all of them with norm pair (2 ó 4), therefore they cannot divide to, whose norm () is odd.

With this we come to the following thing: and they are prime between themselves.

Now, we had the initial equation factorizada of the following form:

Joining these two facts we have that the product of two elements of which they are prime between themselves is equal to a bucket. It forces that each of these elements is he itself a bucket. In particular:

Let's develop now the right part of the latter equality:

Equaling coefficients of of the initial and final expressions we come to the following equality:

A simple analysis of the values of and it takes us that the only possible values are and (let's remember that and there are entire numbers). For obtenemos that and this implies that. And for obtenemos and therefore, that is the looked result.

Do you know any other demonstration of this fact? The comments are yours.