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Math
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2 \times 3
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2^{34}
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a_{i-1}
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Comments
So we are given that x2+y2−x−2kxy=0.
Write x=a2b with bsquare-free. Then
a4b2+y2−a2b−2ka2by=0, so a2∣y2 and a∣y, say y=ac
Then a2b2+c2−b−2kabc=0, so b∣c2 and as b is square-fee in fact b∣c, say c=bd.
Then
a2b+bd2−1−2kabd=0,
so b∣1.
Assume b=−1. Then we have
a2+d2=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 ≡−1(mod4) and the left is ≡+1(mod4)
We conclude b=−1, hence b=+1 and x=a2.
and the proof is completed
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.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
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or__bold__
paragraph 1
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[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
So we are given that x2+y2−x−2kxy=0. Write x=a2b with bsquare-free. Then a4b2+y2−a2b−2ka2by=0, so a2∣y2 and a∣y, say y=ac Then a2b2+c2−b−2kabc=0, so b∣c2 and as b is square-fee in fact b∣c, say c=bd. Then a2b+bd2−1−2kabd=0, so b∣1. Assume b=−1. Then we have a2+d2=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 ≡−1(mod4) and the left is ≡+1(mod4) We conclude b=−1, hence b=+1 and x=a2. and the proof is completed
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Oh that's very nice! Much simpler than the approach I would have taken.
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.
See Vieta Root Jumping .
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Thanks,sir
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Can you please give the main idea?I have been trying it for 3 days.
Also , 8 is a factor of 16 but 8 is not a factor of 4 and 8 is square free.