Читать книгу The Wonders of Arithmetic from Pierre Simon de Fermat - Youri Veniaminovich Kraskov - Страница 9

We will now consider another example of the problem from Fermat's letter-testament, which is formulated there as follows:

There is only one integer square, which increased by two, gives a cube, this square is 25.

When at the suggestion of Fermat, the best English mathematician of the time John Wallis tried to solve it, he was very vexed and forced to acknowledge he could not do it. For more than two centuries it was believed that Leonard Euler received the solution to this problem, but his proof is based on the use of "complex numbers", while we know these are not numbers at all because they do not obey the Basic theorem of arithmetic. And only at the end of the twentieth century André Weil using the Fermat's triangles method still managed to get a proof [17].

It was a big progress because a purely arithmetic method was used here, however, as applied to this problem, it was clearly dragged the ears. Could Fermat solve this problem easier? We will also extract the answer to this question from the cache, what will allow us to reveal this secret of science in the form of the following reconstruction. So, we have the equation p³=q²+2 with the obvious solution p=3, q=5. To prove Fermat's assertion, we suppose that there is another solution P>p=3, Q>q=5, which satisfies the equation

P³=Q²+2 (1)

Since it is obvious that Q>P then let Q=P+δ (2)

Substituting (2) in (1) we obtain: P²(P–1)–2δP–δ²=2 (3)

Here we need just a little bit of “sharpness of mind” to notice that δ>P otherwise equation (3) is impossible. Indeed, if we make a try δ=P then on the left (3) there will be P²(P–4)>2 what is not suitable, therefore there must exist a number δ₁=δ–P. Then substituting δ=P+δ₁ in (3) we obtain

P₂(P–4)–4δ₁P–δ₁²=2 (4)

Now we will certainly notice that δ₁>P otherwise, by the same logic as above, on the left (4) we get P²(P–9)>2 what again does not suitable, then there must exist a number δ₂=δ₁–P and after substituting δ₁=P+δ₂ in (4), we obtain P²(P–9)–6δ₂P–δ₂²=2 (5)

Here one can no longer doubt that this will continue without end. Indeed, by trying δ_i=P each time we get P²(P−K_i)>2. Whatever the number of K_i this equation is impossible because if K_i<P and P>3 then P²(P−K_i)>2 and if K_i≥P then this option is excluded because then P²(P−K_i)≤0

To continue so infinitely is clearly pointless, therefore our initial assumption of the existence of another solutions P>3, Q>5 is false and this Fermat's theorem is proven.

In the book of Singh, which we often mention, this task is given as an example of the “puzzles” that Fermat was “inventing”. But now it turns out that the universal descent method and a simple technique with trying, make this task one of the very effective examples for learning at school.

Along with this proof, students can easily prove yet another theorem from Fermat’s letter-testament, which could be solved only by such a world-famous scientist as Leonard Euler:

There are only two squares that increased by 4, give cubes, these squares will be 4 and 121.

In other words, the equation p³=q²+4 has only two integer solutions.