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anonymous

  • 4 years ago

S is the set of numbers in the form a^2+4ab+b^2 where a and b are integers. If x and y are members of S how can you prove that xy is also in the set? This problem is giving me algebraic meltdown.

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  1. anonymous
    • 4 years ago
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    nice name

  2. anonymous
    • 4 years ago
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    I know that: if x=a^2+4ab+b^2 and y=s^2+4st+t^2 then: xy=a^2 s^2+4 a^2 s t+a^2 t^2+4 a b s^2+16 a b s t+4 a b t^2+b^2 s^2+4 b^2 s t+b^2 t^2 but I can't see how to make this look the correct format.

  3. anonymous
    • 4 years ago
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    Divide the sets into two disjoint subsets A and B such that A consists entirely of (x+y)^2+xy 's and B consists entirely of xy's>It is clear that A +B=S.

  4. anonymous
    • 4 years ago
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    That is both the sets are required to complete the set S

  5. anonymous
    • 4 years ago
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    I am not getting any cogent results, Let say, \[ x = \left(a^2+4 a b+b^2\right) \]and \[ y= \left(m^2+4 m n+n^2\right) \] Then \( xy= a^2 m^2+4 a b m^2+b^2 m^2+4 a^2 m n\) \(+16 a b m n+4 b^2 m n +a^2 n^2+4 a b n^2+b^2 n^2 \) Now this is quite close to \( (a m+b n){}^2 +4 (a n+b m)(a m+b n)+(a n+b m){}^2\) but we will still need \(12 a b m n\).

  6. anonymous
    • 4 years ago
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    I don't follow Aron's solution I'm afraid. Could you possibly make it a bit clearer?

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