Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Some Problems from Diophantus’s Arithmetika II.10. Find two square numbers whose difference is a given number, say 60. If we call one of the unknown squares x2 , then Diophantus’s idea is to name the other one as a variation on that one – by calling it (x + 5)2 . Then the condition of the problem is that (x + 5)2 − x2 = 60, 10x + 25 = 60. Thus the problem has been reduced to a linear equation, which 7 17 35 = and x + 5 = . So the can be solved by simplifying and working in reverse: 10x = 35, x = 10 2 2 2 2 289 49 7 17 = = and differ by 60. squares 2 4 2 4 Diophantus was almost always satisfied with finding one solution of a problem, although he sometimes stated general properties. We can look more closely at this solution technique and find more solutions which would have been acceptable to Diophantus. Instead of x2 and (x + 5)2 , let the two squares be x2 and (x + a)2 . The second degree terms still cancel out, and we are again left with a first degree equation, 2ax + a2 = 60. If a2 is too large, then the solution is negative. So what 60 + a2 60 − a2 we see is that if a is any integer such that 1 ≤ a ≤ 7 then x + a = and x = are 2a 2a 2 2 two positive rational numbers whose squares differ by 60. (Taking a = 6 leads to 8 and 2 as two integral squares whose difference is 60, but Diophantus was satisfied with a more complicated pair.) Questions: If 60 is replaced by another positive integer, does the procedure succeed in solving the new problem? (That is, is every positive integer the difference of two squares – of rational numbers?) Can we find two cubes whose difference is a given number, say 60, either by this method or some other procedure? (Is every positive integer the difference of two cubes – of rational numbers?) A Closer Look at Diophantus’s solution of II.8. A square a2 is given, where a is any positive integer. For illustration purposes, Diophantus took a2 = 16. We wish to express this square as the sum of the squares of two rational numbers. Split it as follows: a2 = x2 + (a2 − x2 ), where x is to be determined. We want a2 − x2 to be a square, so [here is Diophantus’s inspired idea] we will express it as the square of a linear expression involving x and we will choose the constant term in the expression so that the a2 term is matched, i.e., a2 − x2 = (mx − a)2 . (We’ll see shortly why mx − a was chosen instead of mx + a.) Squaring out and simplifying leads us to a linear equation for x, which is the trick we saw in the example above: a2 − x2 = m2 x2 − 2amx + a2 , 2amx = m2 x2 + x2 = (m2 + 1)x2 , 2am = (m2 + 1)x 2am . So no matter what m is, as long as m2 + 1 it is a positive rational number, this value of x will solve the problem for us. since x = 0 isn’t considered to be a number. Thus x = If we use the special value m = 2, which was Diophantus’s choice (i.e., he matched a2 − x2 with (2x − a)2 ), we will obtain x = 4a/5, so that x2 = 16a2 /25, a2 − x2 = 9a2 /25. His conclusion is that every square is the sum of two squares, with the splitting a2 = (3a/5)2 + (4a/5)2 . Notice the Pythagorean triple, 3,4,5! In hindsight, Diophantus could have told his readers that since 32 + 42 = 52 , (3a)2 + (4a)2 = (5a)2 , and so every a2 is the sum of the two squares (3a/5)2 + (4a/5)2 . Generalizing and future results. (a) If we match with a general (mx − a)2 , where m is any natural number, then x = and a2 −x2 turns out to be 2am m2 + 1 (m2 − 1)2 2 2m a . The splitting is a2 = x2 +y 2 wherex = 2 a and y = 2 2 (m + 1) m +1 m2 − 1 a. If you remove the common factor a2 from this relation and then multiply the result by m2 + 1 (m2 + 1)2 , out pops this statement: (2m)2 + (m2 − 1)2 = (m2 + 1)2 . So Diophantus is really rediscovering Plato’s Pythagorean triples!! (b) Diophantus showed that every square is the sum of two squares. When Pierre de Fermat studied the Arithmetika he was fascinated by this fact and tried to extend it in various directions. He wrote that every cube is the sum of three cubes but not the sum of two cubes, every fourth power (bi-quadrate, in his terminology) is the sum of four fourth powers but not the sum of two fourth powers, etc. Part of this collection of statements is the famous “Fermat’s Last Theorem,” which was stated in the 17th century but not proved until the 1990’s.