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BMO 1991

If \(x^2+y^2-x\) is a multiple of \(2xy\) where x and y are integers then prove x is a perfect square.

Note by Lorenc Bushi
2 years, 4 months ago

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So we are given that \[x^2+y^2-x-2kxy=0.\] Write \(x=a^2b\) with \(b\)square-free. Then \[a^4b^2+y^2-a^2b-2ka^2by=0,\] so \(a^2\mid y^2\) and \(a\mid y\), say \(y=ac\) Then \[a^2b^2+c^2-b-2kabc=0,\] so \(b\mid c^2\) and as \(b\) is square-fee in fact \(b\mid c\), say \(c=bd\). Then \[a^2b+bd^2-1-2kabd=0,\] so \(b\mid 1\). Assume \(b=-1\). Then we have \[ a^2+d^2=2kad-1.\] As the right hand side is odd, exactly one of \(a,d\) must be odd, the other even. But then the right hand side is \(\equiv -1\pmod 4\) and the left is \(\equiv +1\pmod 4\) We conclude \(b\ne -1\), hence \[b=+1\] and \[x=a^2.\] and the proof is completed

Refaat M. Sayed - 2 years, 4 months ago

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

Calvin Lin Staff - 2 years, 3 months ago

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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 - 2 years, 4 months ago

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See Vieta Root Jumping .

Calvin Lin Staff - 2 years, 4 months ago

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Thanks,sir

Lorenc Bushi - 2 years, 3 months ago

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

Lorenc Bushi - 2 years, 3 months ago

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

Lorenc Bushi - 2 years, 4 months ago

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