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ParthKohli

  • one year ago

Solving the diophantine equation:\[a^3 + b^3 + c^3 = 100a + 10b + c \]

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  1. perl
    • one year ago
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    hmmm

  2. ParthKohli
    • one year ago
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    Can you help me with that?

  3. perl
    • one year ago
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    find all integer solutions

  4. perl
    • one year ago
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    looks like we can factor something here

  5. perl
    • one year ago
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    did you try to expand (a+b+c)^3

  6. ParthKohli
    • one year ago
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    Whoo.

  7. ParthKohli
    • one year ago
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    \[a^3+3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+b^3+3 b^2 c+3 b c^2+c^3\]

  8. ParthKohli
    • one year ago
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    So do we have to add \(3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+3 b^2 c+3 b c^2\) to both sides?

  9. ParthKohli
    • one year ago
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    \[(a + b + c)^3 = 3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+3 b^2 c+3 b c^2 + 100a + 10b + c \]

  10. perl
    • one year ago
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    oh look at that

  11. perl
    • one year ago
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    one second

  12. perl
    • one year ago
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    (a+b+c)^3 = a^3+3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+b^3+3 b^2 c+3 b c^2+c^3

  13. ParthKohli
    • one year ago
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    How would you solve it now?

  14. perl
    • one year ago
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    not sure

  15. perl
    • one year ago
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    where did you get this question, is it solvable?

  16. perl
    • one year ago
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    well the simple approach is , equate terms

  17. ParthKohli
    • one year ago
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    Yes, a solution is \((1,5,3)\)

  18. perl
    • one year ago
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    |dw:1360413980599:dw|

  19. perl
    • one year ago
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    so if a^2 = 100 , b^2 = 10 , and c^2 = 1 , you have a solution

  20. ParthKohli
    • one year ago
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    I didn't mention that \((a,b,c)\) all must be single-digit numbers.

  21. perl
    • one year ago
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    ok, then you can go through all the cases , 0-9

  22. ParthKohli
    • one year ago
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    I want a purely mathematical solution :-|

  23. ParthKohli
    • one year ago
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    Not by guesses...

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