By Lucien Guillou, Alexis Marin
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Extra resources for A la recherche de la topologie perdue: I Du côté de chez Rohlin, II Le côté de Casson
To show this we just have to work backward: a 2 + b2 4s2 t 2 + (t 2 − s2 )2 (t 2 + s2 )2 c2 , and we are done. The following table shows a few values: t s 2 3 4 4 5 5 6 1 2 1 3 2 4 1 a 2st 4 12 8 24 20 40 12 b t 2 − s2 3 5 15 7 21 9 35 c t 2 + s2 5 13 17 25 29 41 37 There is a lot of geometry behind the Pythagorean triples. For example, one can easily show that the radius of the inscribed circle of a right triangle with integral sides is always an integer (see Problem 2). On the number theoretical side, the components of a 3.
Eqm . Now choose m to If eq is rational, then so is any power (eq )m be the denominator of q (written as a fraction) to get a contradiction to Theorem 1. Remark. Q is dense in R in the sense that, given any real number r, we can ﬁnd a rational number arbitrarily close to r. This is clear if we write r in a decimal representation and chop off √ the tail arbitrarily far away from the decimal point. 414213562 . . 414 13 14 2. “. . There Are No Irrational Numbers at All”—Kronecker √ 1414/1000 . .
Now, if r is the area of a triangle with side lengths a, b, c, then s2 r is the area of a similar triangle with side lengths sa, sb, sc. Thus, we may (and will) assume that a congruent number is a square free natural number. As an example, we immediately know that 6 is a congruent number, since it is the area of a right triangle with side lengths 3, 4, 5. Euler showed that 7 is a congruent number, and Fermat that 1 is not. 4. Eventually, it became clear that 1, 2, 3, 4 are not congruent numbers, while 5, 6, 7 are.
A la recherche de la topologie perdue: I Du côté de chez Rohlin, II Le côté de Casson by Lucien Guillou, Alexis Marin