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 p^>3=q^>2+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

P^>3=Q^>2+2 (1)

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

Substituting (2) in (1) we obtain: P^>2(P–1)–2δP–δ^>2=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 P^>2(P–4)>2 what is not suitable, therefore there must exist a number δ_>1=δ–P. Then substituting δ=P+δ_>1 in (3) we obtain

P_>2(P–4)–4δ_>1P–δ_>1^>2=2 (4)

Now we will certainly notice that δ_>1>P otherwise, by the same logic as above, on the left (4) we get P^>2(P–9)>2 what again does not suitable, then there must exist a number δ_>2=δ_>1–P and after substituting δ_>1=P+δ_>2 in (4), we obtain P^>2(P–9)–6δ_>2P–δ_>2^>2=2 (5)

Here one can no longer doubt that this will continue without end. Indeed, by trying δ_>i=P each time we get P^>2(P−K_>i)>2. Whatever the number of K_>i this equation is impossible because if K_>i

3 then P^>2(P−K_>i)>2 and if K_>i≥P then this option is excluded because then P^>2(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 p^>3=q^>2+4 has only two integer solutions.

We remind that in the Fermat's letter-testament only a special case of this theorem for squares is stated. But also, this simplified version of the task was beyond the power not only of representatives of the highest aristocracy Bachet and Descartes, but even the royal-imperial mathematician Euler.

However, another royal mathematician Lagrange, thanks to the identity found by Euler, still managed to cope with the squares and his proof of only one particular case of FGT is still replicated in almost all textbooks. However, there is no reasonable explanation that the general proof of the FGT for all polygonal numbers obtained by Cauchy in 1815 was simply ignored by the scientific community.