has no positive integer solutions for n > 2.<\/p>\n
Proposed proof. Suppose there were such a solution. Since x\u00a0![\"Unequal\"](\"https:\/\/mathchat.org\/unequal.gif\")
\u00a0y, we may suppose x = y + a, z = y + b, with b > a positive integers. Consider the integer N defined by<\/p>\n
(2) zn -1<\/sup>\u00a0= xn-1<\/sup>\u00a0+ yn-1<\/sup>\u00a0+ N.<\/p>\n Then<\/p>\n xn<\/sup>\u00a0+ yn<\/sup>\u00a0= zn<\/sup>\u00a0= z(xn-1<\/sup>\u00a0+ yn-1<\/sup>\u00a0+ N).<\/p>\n Solving for N yields:<\/p>\n N = [(y+a)n-1<\/sup>\u00a0(a-b) + yn-1<\/sup>\u00a0(-b)]\/(y+b) = [F(y)]\/(y+b) ,<\/p>\n so y+b divides F(y) and<\/p>\n (*) 0 = F(-b) = (a-b)n<\/sup>\u00a0+ (-b)n<\/sup>\u00a00 = (b-a)n<\/sup>\u00a0+ bn<\/sup>\u00a0> 0,<\/p>\n the desired contradiction.<\/p>\n ANSWER.<\/b>\u00a0Luke Gustafson and John Snygg give the best account of the mistake in line (*). It is true that if as expressions in a variable y, y+b divides F(y), then y+b must be a factor of F(y) and F(-b) must be 0. But here, y is a constant, and just because the constant y+b divides the constant F(y), it does not follow that F(-b) = 0.<\/p>\n Elliot Kearsley also points out that the proof, if correct, would apply as well to the false case n = 2. John Robertson also reports that “I have found an error in Wiles’s official published proof of Fermat’s Last Theorem,” but then admits that it is just an insignificant typo.<\/p>\n NEW CHALLENGE.<\/b>\u00a0Which countries in the world have a point such that the shortest line from every other point in the country stays inside the country? (Pretend that the world is round and smooth: ignore mountains and valleys.) Mathematicians call such countries\u00a0starlike<\/i>. Are there any countries such that the shortest line between any two points stays in the country? Mathematicians call such countries\u00a0convex<\/i>.<\/p>\n JOKE.<\/b>\u00a0This week’s winning math joke comes from Eric Brahinsky. Curious scholars at a world mathematics conference are aghast when a man strolls to the lectern leading–a horse!<\/p>\n “Ladies and gentlemen!” he announces, “Meet Dobbin, the world’s only equine mathematical genius! I invite you to submit problems to him, which he will solve with amazing speed!”<\/p>\n The room is in an uproar, but an intrepid voice calls out, “9 times 3!”<\/p>\n Over the din hoofbeats are heard. Twenty-seven of them. The room grows quiet. Then another voice. “Five cubed.” …. And 125 hoofbeats! “6 factorial!”–720 hoofbeats. Applause starts and gradually grows….<\/p>\n “Ladies and gentlemen: these are child’s play. Challenge him! I know you can do better than that!”<\/p>\n “How many ways can 7 distinct objects be partitioned?” snickers one professor. His sneer fades as the horse stamps 877 times. “OK,” snaps another. “How many distinct factors does the square root of Skewes’ number [one of the largest numbers ever to appear in mathematics] have?” With hooves tramping like trillions of troupes of flamenco dancers, Dobbin serenades the astonished gathering with exactly 2.5×10(2×1034-1<\/sup>)<\/sup>+101034<\/sup>\u00a0+1 clicks. “e to the i\u00a0p<\/span>!” Through some mysterious sleight-of-hoof, Dobbin responds with -1 click. Professors assail the heroic horse with the most complex problems in partial differential equations, chaos theory, even…geometric measure theory [Prof. Morgan’s field]. The horse never misses a beat. Dobbin remains unflappable and, evidently, infallible. Applause turns to wild cheering, (human) foot-stomping, cries of “Bravissimo!”<\/p>\n Finally, the beaming presenter holds up one hand. “Ladies and gentlemen: I think we all agree we’ve witnessed a memorable display here. Now we have time for only one more problem, so make it a good one!”<\/p>\n A handful of the world’s top mathematicians huddle together. Pencils are flying, calculators calculating, slide rules sliding, sadistic grins growing.<\/p>\n Amidst all this, a young voice from the back yells, “I got one! I got one!” The gathered scholars turn to see a schoolgirl, maybe 15, with a tattered school text in one hand and the other hand raised and waving madly. Raucous laughter. One slightly annoyed professor says, “Young lady! The world’s top mathematical brains are laboring here. And I suppose you can do better???”<\/p>\n “But I have one he can’t do. I know it. Please? Please?”<\/p>\n The amused presenter says, “In all fairness, she did have hers ready first. Let’s hear it.”<\/p>\n “OK. You’ve got a plane with perpendicular axes and each point identified by its Cartesian coordinates [named after Descartes]. What’s the distance from the origin to (1,0)?”<\/p>\n Gales of laughter. “Come ON! That’s infantile! Go away and let us work!”<\/p>\n “No! Make him answer! He can’t do it!” She strides straight to the front, looks Dobbin straight in the eye and shouts, “Answer it!”<\/p>\n Dobbin twitches his tail a bit but otherwise just stands there. One prof says, “Hey, she’s a sweet kid. He really ought to answer. What’s the deal?” Others join in: “Yeah! What’s the matter, horse?”<\/p>\n Dobbin’s tail twitches more, and his mane begins to toss. Nostrils flare. A slight lather can be seen on his flanks. But the assemblage listens in vain for that single click.<\/p>\n “Well, obviously,” says the presenter, “such a silly trifle isn’t worth his time.” But by now a chant of “ANSWER IT! ANSWER IT!” has risen. The horse’s eyes are wild with fear. Finally, Dobbin rears up, neighs in terror, and bolts from the room–his humiliated owner running after him to calls of “Fraud! Charlatan!”–and worse.<\/p>\n Back in the room, the hubbub eventually settles. One mathematician approaches the girl who had asked the question that brought Dobbin down. “It was such an easy question. How did you know?”<\/p>\n “Oh, that’s simple. Everyone knows you can’t put Descartes before the horse.”<\/p>\n Send answers, comments, and new questions by email to:<\/p>\n Frank.Morgan@williams.edu<\/a>, to be eligible for\u00a0Flatland\u00a0<\/i>and other book awards. Winning answers will appear in the next Math Chat. Math Chat appears on the first and third Thursdays of each month. Prof. Morgan’s homepage is at\u00a0www.williams.edu\/Mathematics\/fmorgan<\/a>.<\/p>\n <\/p>\n Copyright 1999, Frank Morgan.<\/p>\n","protected":false},"excerpt":{"rendered":" March 4, 1999 This Math Chat reveals the flaw in the short proof of Fermat’s Last Theorem, provides a new challenge question, and gives this week’s somewhat lengthy math joke. OLD CHALLENGE.\u00a0Critique the following short proof of Fermat’s Last Theorem sent in by reader Rob Connelly. (In perhaps the biggest mathematics news of the […]<\/p>\n","protected":false},"author":2965,"featured_media":0,"parent":3459,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_acf_changed":false,"footnotes":""},"class_list":["post-3622","page","type-page","status-publish","hentry"],"acf":[],"_links":{"self":[{"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/pages\/3622","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/users\/2965"}],"replies":[{"embeddable":true,"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/comments?post=3622"}],"version-history":[{"count":1,"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/pages\/3622\/revisions"}],"predecessor-version":[{"id":3623,"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/pages\/3622\/revisions\/3623"}],"up":[{"embeddable":true,"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/pages\/3459"}],"wp:attachment":[{"href":"https:\/\/sites.williams.edu\/Morgan\/wp-json\/wp\/v2\/media?parent=3622"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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