|
||
|
Title: An IMO question: (4ab-1) | (4a^2 - 1)^2 Post by wonderful on May 29th, 2008, 6:39pm Let a and b be positive integers. Show that if (4ab-1) divides (4a^2 - 1)^2 , then a=b. Have A Great Day! Source: IMO |
||
|
Title: Re: a, b postive integers Post by ecoist on Jun 6th, 2008, 8:31pm I bet everyone interested has observed this. 4ab-1 divides (a-b)2 and therefore also divides (4b2-1)2. |
||
|
Title: Re: a, b postive integers Post by Aryabhatta on Jun 8th, 2008, 10:57am We can also show that there if 4ab-1 divides (4a^2-1)^2 and (b =/= a), then there must be some b' < a such that 4ab'-1 divides (4a^2-1)^2, which combined with your observation would give us an infinite decreasing sequence of natural numbers... To show that there is a b' < a: We can show that (4a^2-1)^2/(4ab-1) must be of the form 4ab' - 1 with 0 < b' < a, if b > a. |
||
|
Title: Re: a, b postive integers Post by ecoist on Jun 9th, 2008, 2:43pm Wow, Aryabhatta, you can squeeze blood from a turnip! I thought my observation was hardly worth posting. |
||
|
Title: Re: a, b postive integers Post by wonderful on Jun 10th, 2008, 5:38pm This question may have sevaral solution. For those who are interested, attached is a nice solution. Have A Great Day! |
||
|
Title: Re: a, b postive integers Post by ecoist on Jun 11th, 2008, 4:48pm Pray tell, what's the "4" for? Seems irrelevant to both proofs. Sorry, still fascinated by both the problem and Aryabhatta's proof. (And isn't it time to correct the spelling in the title?) |
||
|
Title: Re: An IMO question: (4ab-1) | (4a^2 - 1)^2 Post by Aryabhatta on Jun 18th, 2008, 7:21pm Sorry, i lost track of this. ecoist, i guess i just got lucky. If had posted my observation before you did, i am sure you would have completed the proof... |
||
|
Title: Re: An IMO question: (4ab-1) | (4a^2 - 1)^2 Post by ecoist on Jun 18th, 2008, 8:32pm Like Tiger Woods winning his 14th major golf tournament, talented people are often lucky, Aryabhatta! Even though I considered infinite descent, your simple solution never occured to me. |
||
|
Title: Re: An IMO question: (4ab-1) | (4a^2 - 1)^2 Post by Hippo on Jun 18th, 2008, 10:22pm I didn't think about it much, but I don't understand the Aryabatta's proof: ... there is no a-b symmetry and b>a -> b'<a. How do you gain b''? |
||
|
Title: Re: An IMO question: (4ab-1) | (4a^2 - 1)^2 Post by ecoist on Jun 19th, 2008, 10:00am There is symmetry, Hippo, because 4ab-1 also divides (4b2-1)2 (see my post)! Hence Aryabhatta can assume that b>a if a=/=b. Then he shows that there exists b'<a such that the pair {a,b'} satisfies the same conditions (including b>a) as the pair {a,b}, with b' as the new "a" and a as the new "b". It now follows that there is a b''<b', which pair {b'',b'} also satisfies the same conditions with b'' as the new "a" and b' as the new "b". |
||
|
Title: Re: An IMO question: (4ab-1) | (4a^2 - 1)^2 Post by Hippo on Jun 19th, 2008, 3:43pm OK ... thanks for explanation. P.S.: Now I get the time to look at it ... nice ;) ... symmetry 16a2b2(a-b)2 = (4ab-1)(4ab3 + b2 - 8a2b2) + (4a2-1)2 |
||
|
Powered by YaBB 1 Gold - SP 1.4! Forum software copyright © 2000-2004 Yet another Bulletin Board |