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LollyLau

  • 5 years ago

Series of trial-and-error questions: Find all the unknowns for: 1. x^y=y^x 2. x^y=y^x+1 where all the unknowns are integers.

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  1. LollyLau
    • 5 years ago
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    ?????????!!!!!!!!

  2. anonymous
    • 5 years ago
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    i guess \[y=x\] might work for first one

  3. LollyLau
    • 5 years ago
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    could be...

  4. LollyLau
    • 5 years ago
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    let's say they aren't, some actual answers please...

  5. anonymous
    • 5 years ago
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    what is wrong with \[3^3=3^3\]?

  6. anonymous
    • 5 years ago
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    2 and 4 works for the first one.

  7. LollyLau
    • 5 years ago
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    Nothing! Pratu got it right :) (Well you too...)

  8. Mr.Math
    • 5 years ago
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    Sorry, I thought it was a system of two equations.

  9. LollyLau
    • 5 years ago
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    oh.

  10. anonymous
    • 5 years ago
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    ooooooooooooooh. i was wondering...

  11. Mr.Math
    • 5 years ago
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    I agree with satellite on the first one, the solution of the first one is all integers (x,y) such that x=y. For the second one, we have (x,y)=(2,1) is a solution.

  12. Mr.Math
    • 5 years ago
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    (3,2) is another solution.

  13. anonymous
    • 5 years ago
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    pratu had \[2^4=4^2\] so that works as well. my guess is there are no others, but that is a guess

  14. LollyLau
    • 5 years ago
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    There are no others.

  15. LollyLau
    • 5 years ago
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    no proof tho. just know that for SURE :)

  16. LollyLau
    • 5 years ago
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    Mr. Math, you're missing one more answer. (which i know)

  17. LollyLau
    • 5 years ago
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    (solution)

  18. LollyLau
    • 5 years ago
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    ans. in half a min.

  19. Mr.Math
    • 5 years ago
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    (1,0).

  20. LollyLau
    • 5 years ago
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    YAY!

  21. Mr.Math
    • 5 years ago
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    Actually any solution in the form (x,0), x is a positive integer.

  22. LollyLau
    • 5 years ago
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    Can we do this for +2, +3, +4 etc.

  23. Mr.Math
    • 5 years ago
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    Yes.

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