BMO 1991

If x2+y2xx^2+y^2-x is a multiple of 2xy2xy where x and y are integers then prove x is a perfect square.

Note by Lorenc Bushi
5 years, 10 months ago

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Comments

So we are given that x2+y2x2kxy=0.x^2+y^2-x-2kxy=0. Write x=a2bx=a^2b with bbsquare-free. Then a4b2+y2a2b2ka2by=0,a^4b^2+y^2-a^2b-2ka^2by=0, so a2y2a^2\mid y^2 and aya\mid y, say y=acy=ac Then a2b2+c2b2kabc=0,a^2b^2+c^2-b-2kabc=0, so bc2b\mid c^2 and as bb is square-fee in fact bcb\mid c, say c=bdc=bd. Then a2b+bd212kabd=0,a^2b+bd^2-1-2kabd=0, so b1b\mid 1. Assume b=1b=-1. Then we have a2+d2=2kad1. a^2+d^2=2kad-1. As the right hand side is odd, exactly one of a,da,d must be odd, the other even. But then the right hand side is 1(mod4)\equiv -1\pmod 4 and the left is +1(mod4)\equiv +1\pmod 4 We conclude b1b\ne -1, hence b=+1b=+1 and x=a2.x=a^2. and the proof is completed

Refaat M. Sayed - 5 years, 10 months ago

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Oh that's very nice! Much simpler than the approach I would have taken.

Calvin Lin Staff - 5 years, 9 months ago

Thanks for taking your time to it, but do you mind giving some clarification.How do you know that x is not prine or a product of primes per example 5 ,6,15,13 etc and cannot be written as you stated in your solution. Forgive me if im missing some important clue. Also i forgot to state that x and y are POSITIVE INTEGERS. Disregard this reply if you were based heavily on this conditiom.

Lorenc Bushi - 5 years, 10 months ago

See Vieta Root Jumping .

Calvin Lin Staff - 5 years, 9 months ago

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Thanks,sir

Lorenc Bushi - 5 years, 9 months ago

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Can you please give the main idea?I have been trying it for 3 days.

Lorenc Bushi - 5 years, 9 months ago

Also , 8 is a factor of 16 but 8 is not a factor of 4 and 8 is square free.

Lorenc Bushi - 5 years, 10 months ago
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