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This problem is from ( BMO Exam1991 ). I tried to solve but it was difficult.

The problem is:

If $ x^{2} + y^{2} - x $ is a multiple of $ 2xy $ where $x$ & $y$ are integers, prove that $x $ is a perfect square.

Any help or hint please.

Bart Michels
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user257567
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2 Answers2

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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 $$x=a^2.$$

Hagen von Eitzen
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  • Nice proof sir. Thanks – user257567 Aug 19 '15 at 13:09
  • @user257567 , if you think this is the best answer to your question, please accept it by clicking on the check sign. I noticed that you've done that on other question, so I thought you might not know about this. – Jean Aug 19 '15 at 14:04
  • @JeanYasir sorry. But I couldn't understand you.. what you want to say in your comment? – user257567 Aug 19 '15 at 17:45
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Another way：

$x^2+y^2-x\equiv x(x-1)\mod y\Rightarrow y|x(x-1)$, which means $y=y_1y_2$, $y_1|x$ and $y_2|(x-1)$, and $(y_1,y_2)=1$

$x^2+y^2-x\equiv y^2\mod x\Rightarrow x|y^2=y_1^2y_2^2\Rightarrow x|y_1^2$

plug into $x=my_1$, where $m|y_1$, $y_1^2|x^2+y^2-x=y_1(m^2y_1-m+y_1y_2^2)\Rightarrow y_1|(m^2y_1-m+y_1y_2^2)\Leftrightarrow y_1|m$ so $m=y_1$, $x=y_1^2=(x,y)^2$

chenyuandong
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