semester we used the following axiom equivalent to mathematical induction or

Semester we used the following axiom equivalent to

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semester we used the following axiom, equivalentto mathematical induction, or the principle of min-imization (also called the well-ordering principle).Axiom.There is no infinite decreasing sequence ofpositive integers.Euclid used this principle frequently in hisEle-ments. Fermat used it in number theory, too. Fer-mat would take a particular Diophantine equation,suppose that there was one solution, and look fora way to produce a smaller solution from it. If thisworked without fail, then he would have an infinitedecreasing sequence of solutions and could concludethat no solution could exist. On the other hand, ifthe descent stopped with an actual solution, revers-ing the process would lead to more solutions.We’ll look at Fermat’s proof thatx4+y4=z4hasno positive integral solutions. Actually, he provedsomething stronger, that the Diophantine equationx4+y4=z2has no solutions.Theorem.There are no positive integral solutionsofx4+y4=z2.Proof: Suppose thatx, y, zis a solution.First, we want to reduce to the case wherexandyare relatively prime.Letdbe the great-est common divisor of them, and letx1=x/dandy1=y/d. Thend4(x41+y41) =z2.Therefore,d4|z2, sod2|z. Letz1=z/d2. Thenx41+y41=z21.We now have a smaller solution wherex1andy1are relatively prime.Now we may suppose thatx, y, zis a solutionwhere (x, y) = 1.Following Fermat, we need tofind a smaller solution, and in this case that meansa solution with a smaller value forz.Note that (x2)2+ (y2)2=z2, so (x2, y2, z) isa primitive Pythagorean triple.It turns out thatusing the alternate parameterizations of primitivePythagorean triples works better here than the pa-rameterization we found. We’ll takex2odd, then(x2, y2, z) = (u2-v2,2uv, u2+v2)where whereuandvare positive relatively primeintegers withu > v.Note thatx2=u2-v2,sox2+v2=u2.Therefore, (x, v, u) is anotherPythagorean triple, and it’s primitive sinceuandvare relatively prime. Therefore,(x, v, u) = (s2-t2,2st, s2+t2)where wheresandtare positive relatively primeintegers withs > t.Now,y2= 2uv.Sinceuandvare relativelyprime, anduis odd (because (x, v, u) is a primitivePythagorean triple), thereforeuand 2vare rela-tively prime. Buty2is their product, so each ofuand 2vis a square. Letu=z22and2v=c2,and sincecis even, letc= 2dso thatv= 2d2.Sincev= 2st, thereforest=v/2 =d2. 4
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  • Fall '14
  • D.Joyce
  • Math, Number Theory, Prime number, Euclidean algorithm, Pythagorean triple, pythagorean triples, Fermat's Last Theorem

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