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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''.

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    $\begingroup$ For Euler's determination of the rational points on $y^2=x^3+1$, see this note by Joseph Oesterlé :docs.google.com/… $\endgroup$ Sep 15 '13 at 16:41
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    $\begingroup$ hardly historical, but observing that this problem is equivalent to finding integral points on an elliptic curve of rank 1, it would be surprising if any decent general methods were available at that time for problems like this one. now a little curious how the more elementary methods might translate into contemporary geometric language. lmfdb.org/EllipticCurve/Q/1728/o/3 $\endgroup$
    – pupshaw
    Jul 5 at 14:03
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    $\begingroup$ @pupshaw: My answer from earlier today only uses methods available in Fermat’s time. Would love to see you translate it into contemporary math! $\endgroup$ Jul 5 at 17:55
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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.

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    $\begingroup$ Yes the fact that Fermat claimed to have a proof sounds familiar.(actually i have one too,but the comment space has too little characters to analyse it!)So, i supppose that all people mentioning that ''Fermat had always a proof for his statements except once'' (Fermat's Last Theorem) should say ''except twice''? $\endgroup$ Sep 15 '13 at 22:34
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    $\begingroup$ @KonstantinosGaitanas, Fermat didn't have a proof for his statement that $2^{2^n}+1$ is always prime. $\endgroup$ Sep 15 '13 at 23:47
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    $\begingroup$ Yes you are right.Three times then. $\endgroup$ Sep 16 '13 at 10:38
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    $\begingroup$ To be accurate, Fermat never claimed to have proven that $2^{2^n}+1$ is always prime — rather he said that after much research he believed it to be true, and [on several occasions] asked others to help him prove it. All of the theorems which he claimed to have actually proven were indeed verified by later mathematicians. $\endgroup$ Oct 20 '13 at 1:47
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    $\begingroup$ I think we are too quick to say that Fermat had no proof of his claim about his equation. If you read Fermat's correspondance, you see there was a reason he rarely wrote down his proof: essentially no one would read them. When he talks to Pascal, for example, Pascal is eager to talk about probability, physics, and other subjects but show little appetite for reading Fermat's complicated number-theory proofs. Fermat was simply too much in advance of his time. His best interlocutors would have been he Bernoulli and Euler, but they were born after his death. $\endgroup$
    – Joël
    Jul 30 '15 at 14:24
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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.

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    $\begingroup$ Observe that \begin{align} y^2 + 2 &= x^3 \\ y^2 + 3 &= x^3+1\\ &=(x+1)(x^2-x+1) \\ \frac{y^2+3}{4} &= \frac{x+1}{4} \cdot \biggl(\frac{x+1}{4}4(x-2)+3\biggr). \end{align} It is relatively easy to show that $(y^2+3)/4$ and $(x+1)/4$ are both odd, so we have the form $a^2+3b^2$ on the left-hand side, and two factors which should be of the same form on the right. Fermat was playing around with the form $a^2 + 3b^2$ at the same time as he claimed his two “elliptic curve” results. Maybe he used something like this to find a descent path? Apologies if you (Franz) already tried this road… $\endgroup$ Oct 20 '13 at 2:05
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    $\begingroup$ Note also that a similar method would be applicable to his second "elliptic curve theorem", as \begin{align} y^2+4 &= x^3 \\ y^2+3 &= x^3-1 \\ &= (x-1)(x^2+x+1) \end{align} which puts us in the same position as the other example. If Fermat had a method for one, he almost certainly was able to apply it almost immediately to the other. $\endgroup$ Oct 20 '13 at 2:11
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    $\begingroup$ p.s. If you want to discuss this further, feel free to email me! kieren at alumni.rice.edu $\endgroup$ Oct 20 '13 at 2:24
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    $\begingroup$ Another interesting observation is that any number of the form $a^2+3b^2$ is the sum of three triangular numbers whose "roots" sum to zero. Fermat was a master with polygonal numbers in general — and triangular numbers in particular — so I've always felt that was part of his method. $\endgroup$ Oct 21 '13 at 12:56
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    $\begingroup$ I noticed that, in the first case, if we write $x+1=a$ and $x^2-x+1=b$, then $y^2+3=ab$ and $a+b=x^2+2$, which is of the same form as the $y$ side of the original equation, i.e. $y^2+2=x^3$. Might be coincidence, but still seemed noteworthy. $\endgroup$ Oct 24 '13 at 0:28
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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.

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    $\begingroup$ "I personally believe this finally puts to rest any questions of whether Fermat had a proof of these two claims." Well, it sure puts to rest any questions of whether Fermat could have had a proof of the two claims. $\endgroup$ Feb 25 at 1:54
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    $\begingroup$ @GerryMyerson: That’s what I meant. =) Edited. $\endgroup$ Feb 25 at 2:28
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    $\begingroup$ @Gerry: I disagree. If you remove everything from the proof that Fermat would not have used (complex numbers, negative numbers, extensive algebraic manipulations) you end up with something that is very difficult to follow. The algebra contained in the proofs Fermat has left is much less advanced. Of course we can also write this up using only Euclidean mathematics - but this does not prove anything. $\endgroup$ Feb 25 at 15:54
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    $\begingroup$ Fermat “would not have used” negative numbers? What an odd (and, I think, indefensible) claim… He used them all the time — in fact, he explicitly described how to get around situations where solutions gave negative numbers (adequality, etc.).And the “complex numbers” in that proof are very easily replaced by equivalent algebra in the real numbers. $\endgroup$ Feb 25 at 19:20
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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 $(x,y)=(3,\pm 5)$. $\blacksquare$

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    $\begingroup$ Lemma and Corollary (in the case $a^2+3b^2$) due to Stan Dolan. Adjustments (for $a^2+2b^2$) and Theorem due to multiple Gazette readers. $\endgroup$ Jul 5 at 13:36
  • $\begingroup$ Note the use of Fermat’s method of descent (“minimal counterexample”) in the Lemma. $\endgroup$ Jul 5 at 13:36
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    $\begingroup$ I presume this is a "cleaned up" version of the proof mentioned in your other answer, shorn of all the objectionable elements? $\endgroup$ Jul 6 at 6:26
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    $\begingroup$ @DavidRoberts: Yes. I wrote it out in full here because it was too big to fit in the margin… ;) $\endgroup$ Jul 16 at 16:54
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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

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    $\begingroup$ There may be more than one way of writing numbers in the form $a^2+2b^2$. So the last para doesn't seem clear to me. $\endgroup$
    – Lucia
    Jul 29 '15 at 17:46
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    $\begingroup$ There is more than one way for non-primes, but b^2 = 1 no matter how you write it. $\endgroup$
    – Bob
    Jul 29 '15 at 17:52
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    $\begingroup$ This proof was advocated by Weil in his book "number theory: an approach through history"; Weil of course also identifies the lemma that is missing from the "proof" above: one has to count the number of representations of integers by the form $x^2 + 2y^2$, and this is exactly the point where things get technical. $\endgroup$ Jul 30 '15 at 18:03
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    $\begingroup$ Franz - don't know what you mean, y^2=1, count them there is one representation. Never read the book. $\endgroup$
    – Bob
    Jul 30 '15 at 20:45
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    $\begingroup$ @Bob Can you prove there is only one representation? $\endgroup$
    – Wojowu
    Jul 30 '15 at 22:05

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