Mancosu - Philosophy of Mathematical Practice (Oxford, 2008).pdf

‘there is no ontology here’ 375who, having quoted Kronecker’s dictum on page 1 of his yellow Vorlesungenüber Zahlentheorie (1950), added to the index of names at the book’s end underthe letter L the entry ‘Lieber Gott p. 1.’ (26 May 1999 post to the list HistoriaMatematica archived at )Kronecker was not serious about the theology of numbers. He was serious aboutreplacing irrational numbers with the ‘pure’ arithmetic of integer polynomials.14.1.3 One Diophantine equationConsider this Diophantine equation:Y 2 = 3X + 2 (14.1)Calculation modulo 3 will show it has no integer solutions. Say that integersa, b are congruent modulo 3, and writea ≡ b (mod 3)if and only if the difference a − b is divisible by 3. The key here is thatcongruent numbers have congruent sums and products.Theorem 5. Suppose a ≡ b and c ≡ d(mod 3). Then(a + c) ≡ (b + d) and (a · c) ≡ (b · d) (mod 3)Proof. Suppose 3 divides both a − b and c − d. The claim follows because(a + c) − (b + d) = (a − b) + (c − d)(a · c) − (b · d) = (a − b) · c + b · (c − d) □If Equation (14.1) had an integer solution X = a, Y = b then the sideswould also be congruent modulo 3b 2 ≡ 3a + 2 ≡ 2 (mod 3)But b is congruent to one of {0, 1, 2} modulo 3. By Theorem 5 the squareb 2 is congruent to the square of one of these. And none of these squares is 2modulo 3:0 2 ≡ 0 1 2 ≡ 1 2 2 ≡ 4 ≡ 1 (mod 3)So a, b cannot be an integer solution to Equation (14.1).For future reference notice that Equation (14.1) does have solutions moduloother integers. For example X = 2, Y = 1 is a solution modulo 7 since1 2 ≡ 1 ≡ 8 ≡ 3 · 2 + 2 (mod 7)