# 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, 11 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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

- 2 years, 11 months ago

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

Staff - 2 years, 11 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.

- 2 years, 11 months ago

Staff - 2 years, 11 months ago

Thanks,sir

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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

- 2 years, 11 months ago