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In his letter to Carcavi (August 1659), Fermat mentions the following challenge

There is no number, one less than a multiple of $3$, composed of a square and the triple of another square.

He says that he has solved it using infinite descent. The next problem that he proposes in the same letter is also solved by him using infinite descent (which was discovered in his copy of Diophantus Arithematica).

However, this question is not particularly clear to me, as to what is being asked to prove here. Can anyone please restate it using modern notations and provide a proof using Infinite Descent?

Thanks in advance.

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Not entirely sure...but it sounds like the question is "Show that, if $n=3k-1$ then $n$ can not be expressed as $a^2+3b^2$ for integers $a,b$".

But, that is obviously true (as it would imply that $-1$ was a square $\pmod 3$ which it is not).

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  • $\begingroup$ But 3.5 centuries ago, it was not so obvious, I guess. $\endgroup$ – peter.petrov Mar 16 '16 at 13:20
  • $\begingroup$ @peter.petrov To Fermat? Maybe...but infinite descent is, well, infinitely more subtle than a simple congruence argument. It is true that we're a lot more used to residue arguments so....maybe. $\endgroup$ – lulu Mar 16 '16 at 13:27
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This letter to Carcavi is not preserved; what we do have is a copy of the main part of the letter made, I think, by Huygens. I believe that Huygens misread Fermat's statement, since Fermat would not have included such a trivial statement in a letter that may be seen as his number theoretical testament. What Fermat probably wrote is that numbers divisible by some prime of the form $3n-1$ cannot be written in the form $x^2 + 3y^2$, and this may be proved by infinite descent in the usual way (see https://mathoverflow.net/questions/88539/sums-of-rational-squares).

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  • $\begingroup$ Oh. Thanks for the info. Can you please post the link to your proof? $\endgroup$ – Henry Mar 21 '16 at 16:06
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Infinite Descent is not required.

Well first note that all squares have a form of $3m$ or $3m+1$.

It can be observed that then

$$a^2 + 3b^2 \neq 3k -1$$

See that the RHS is of the form $3k_1+2$.

$3b^2$ is a multiple of 3 so $a^2$ must be of the form $3m+2$ which is not possible.

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