## Posts Tagged ‘**quality**’

## Fermat’s last theorem

Fermat’s last theorem has a long and exciting history. Which everyone knows, so I’ll not mention it here.^{1} What I suddenly find to be remarkable though, is the *very first event*. The fact that *Fermat scribbled it in a margin of Diophantus’s Arithmetica.* That Pierre de Fermat, in France in 1637, was reading an ancient book written by a Greek in the 3rd century. That he was reading it in such a manner that the book’s asking how to split a square into two squares should impel him to not only investigate the question of how to split a nth power into two nth powers, for all n, but to also do it until he believed he had a *truly marvelous proof*.

When was the last time you made margin notes in a book?

Off topic: The book only answers the question for 16(=4²). Wikipedia has pictures of the relevant page for a 1621 edition, and the 1670 edition that contains Fermat’s notes. (Fermat died in 1665.) I’m not sure I’ve deciphered the Latin correctly (the Greek is right out), but what it says is the following.

[BTW, in case you have been thinking so far and have the objection that 16 *cannot* be written as the sum of two squares, I should point that for Diophantus, “number” apparently meant “positive rational number”, there were no other kinds of numbers. Negative and irrational numbers were “useless”, “meaningless”, and “absurd”.]

Suppose one of the two squares that add up to 16 is Q=N². [“Q” because it is a square, “quadratum”.] The other square is 16-Q. If the other square is (2N-4)²=4Q+16-16N, [um, why should it be?] then we get 16-Q=4Q+16-16N so 5Q=16N, or N=16/5 and Q=N²=256/25 (which is misprinted as 256/52 in the 1670 edition), and the other square is 144/25, which add up to 400/25=16. So the (*an*) answer is that 16 = (16/5)² + (12/5)².

You might notice this is not really an answer; all that Diophantus has done is take 3²+4²=5² and multiplied it appropriately to make two “squares” add up to 16. We could do the same for any square, e.g. for 49=7², we could write (7×3)²+(7×4)²=(7×5)², then divide out by 5² to say (21/5)²+(28/5)²=49. For any x, we could take any a and b such that a²+b²=1 (e.g. 3/5 and 4/5) and write x²=(ax)²+(bx)².

↑**1.** I found today (2008-11-27) an anecdote. The setting is this: it was April 1994. Andrew Wiles had first announced his proof in June the previous year, and sent it off to a journal, but a hole had been found. It seemed at first it would take only a few hours, then weeks, to fix it, but months had dragged on without success. And on April 3 1994, Gian-Carlo Rota sent out an email announcing that Noam Elkies had found a counterexample to Fermat’s last theorem! So it seemed that the hole was unfixable after all. There was some disappointment all around before it was realised that the email was an April Fool’s joke, that had somehow got incorrectly dated :-) I found it on Wikibooks, but see Lance Fortnow’s blog post for the email.