Fermat's proof for $x^3-y^2=2$

Question

Fermat proved that $x^3-y^2=2$ has only one solution $(x,y)=(3,5)$.

After some search, I only found proofs using factorization over the ring $Z[\sqrt{-2}]$.

My question is:

Is this Fermat's original proof? If not, where can I find it?

Thank you for viewing.

Note: I am not expecting to find Fermat's handwritings because they may not exist. I was hoping to find a proof that would look more ''Fermatian''.

Answer 1

Fermat never gave a proof, only announced he had one (sounds familiar?). Euler did give a proof, which was flawed, see Franz Lemmermeyer's lecture notes, or see page 4 of David Cox's introduction.

For a discussion why a proof along the lines set out by Fermat is unlikely to work, see this MO posting.

---- trivia ----

As a curiosity, I looked up Fermat's original text (reproduced below from his collected works), written in the margin of the Arithmetica of Diophantus:

Can one find in whole numbers a square different from 25 that, when increased by 2, becomes a cube? This would seem at first to be difficult to discuss; and yet, I can proof by a rigorous demonstration that 25 is the only integer square that is less than a cube by two units. For rationals, the method of Bachet would provide an infinity of such squares, but the theory of integer numbers, which is very beautiful and subtle, was not known previously, neither by Bachet, nor by any author whose work I have read.

Answer 2

Fermat did not prove this result; he claimed that the only solution is the obvious one and conjectured (in words that seem to suggest he knew how to prove it, but without explicitly saying so) that this can be proved by descent. I am sure that Fermat, if he really believed to have a proof (in my opinion he did not), was mistaken.

I am not aware of any proofs based on Fermat's techniques alone, and I have often tried to find one myself - so far without success.

Answer 3

A completely elementary proof can be found on page 561 of the Nov 2012 edition of The Mathematical Gazette, where a descent mechanism first used by Stan Dolan in the March 2012 edition is adapted (as per his challenge to the “interested reader”) to solve both of Fermat’s “elliptic curve” theorems. The method uses math which was clearly available in Fermat’s time, and in particular to Fermat himself.

I personally believe this finally puts to rest any questions of whether Fermat could have had a proof of these two claims.

Answer 4

Lemma. Let $a$ and $b$ be coprime integers, and let $m$ and $n$ be positive integers such that $a^2+2b^2=mn$. Then there are coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$. Furthermore, for any such choice of $r$ and $s$, there are coprime integers $t$ and $u$ such that $a=rt-2su$, $b=ru+st$, and $n=t^2+2u^2$ divides $bt-au$.

Proof. Assume the theorem is false, and let $m$ be a minimal counterexample. Evidently $m > 1$ since the theorem is trivially true for $m=1$.

Note that $b$ is coprime to $m$. Let $A$ be an integer such that $Ab \equiv a\!\pmod{m}$, chosen so that $\tfrac{-m}{2} < A \le \tfrac{m}{2}$. Then $A^2+2 = lm$ for some positive integer $l < m$. Clearly $l$ cannot be a smaller counterexample than $m$, and so there exist coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$.

Let $t = \tfrac{ar+2bs}{m}$ and $u=\tfrac{br-as}{m}$. Direct calculation confirms the equations for $a$, $b$, and $n$. From $n=t^2+2u^2$, we deduce that $t$ is an integer because $u$ is an integer, and $t$ and $u$ are coprime because $\gcd(t,u)$ divides both $a$ and $b$. Finally, note that $n$ divides $bt-au=sn$.

Hence $m$ is not a counterexample, contradicting the original assumption. $\blacksquare$

Corollary. Let $a$ and $b$ be coprime integers with $m$ an integer such that $m^3=a^2+2b^2$. Then there are coprime integers $r$ and $s$ such that $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$.

Proof. Evidently $m$ is odd since $a^2+2b^2$ is at most singly even. And $a$ and $m$ must be coprime. Using the theorem, we have $m=r^2+2s^2$ and $m^2=t^2+2u^2$. Then $m$ divides $a(ur-ts)=t(br-as)-r(bt-au)$, and therefore $m \mid (ur-ts)$. The lemma can then be reapplied with $a$ and $b$ replaced by $t$ and $u$. Repeating the process, we eventually obtain integers $p$ and $q$ such that $p^2+2q^2=1$. The only solution is $q=0$ and $p=\pm1$. Ascending the path back to $a$ and $b$ (reversing signs along the way, if necessary) yields $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$, as claimed. $\blacksquare$

Theorem. The Diophantine equation $X^3 = Y^2+2$ has only one integer solution, namely $(x,y) = (3, \pm 5)$.

Proof. Evidently $y$ and $2$ are coprime. By the corollary, we must have $b=1=s(3r^2-2s^2)$ for integers $r$ and $s$. The only solutions are $(r,s)=(\pm 1,1)$. Hence $a=y=r(r^2-6s^2)=\pm 5$, so $(x,y)=(3,\pm 5)$. $\blacksquare$

Answer 5

Here is how Fermat probably did it (it is how I did it - not all of the steps were needed but I have to believe this was close to Fermat's thought process).

Any prime of the form $8n+1$ or $8n+3$ can be written in the form $a^2 +2b^2$. This is proved with descent techniques once realizes that $-2$ and $1$ are squares mod $8n+1$ or $8n+3$ and hence setting $a^2=-2$ and $b^2 = 1$ gets the result of $0$ (mod $8n+1$ or $8n+3$) for $a^2+2b^2$, which means our prime divides the result. Any prime of the form $8n+5$ or $8n+7$ cannot be.

Point two is that combinations of squares with common shapes when multiplied by each other retain their shape. Let $x = a^2 + Sb^2$, and $y = c^2 + Sd^2$. $xy = (ac+Sbd)^2 + S(ad-bc)^2 = (ac-Sbd)^2 + S(ad+bc)^2$

Point three is that if $y$ is even $y^2 + 2$ is even as is $x^3$. Dividing both sides by $2$ would make the left hand side odd and right hand side even so both $y$ and $x$ are odd.

Point four is that if a non-prime is of the form $a^2 + 2b^2$ then all its prime factors must be of the form $8n+1$ or $8n+3$, or the factor must be a square.

Point five is to observe that $y^2 + 2$ is of the form $a^2 + 2b^2$ with $a=y$ and $b=1$. Combining this with four and one means there are no squares of the form $8n+5$ or $8n+7$ since $b$ would be equal to that square, not $1$.

So now we expand upon point three to make the proof. $x$ is of the form $a^2 + 2b^2$. $x^3$ can be written as $(a^3-3Sab^2)^2 + S(3a^2b-Sb^3)^2$. Letting $S=2$ we see that the expression $(3a^2b-2b^3)^2$ must be equal to $1$. Hence $b^2 \cdot (3a^2-2b^2)^2 =1$. Using positive integers we see $b=a=1$ is the only solution. Hence $x =1^2 + 2*1^2 = 3$ is the only possibility and $5^2 + 2 =3^3$ is the only solution