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  1. MHB Craftsman
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    #1
    Find the pairs of nonnegative integers, $(m,n)$, which obey the equality:

    \[(m-n)^2(n^2-m) = 4m^2n\]

    So far, I havenīt found a single pair, but I cannot prove, that the set of solutions is empty.
    Perhaps, someone can help me to crack this nut?

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    #2
    Quote Originally Posted by lfdahl View Post
    Find the pairs of nonnegative integers, $(m,n)$, which obey the equality:

    \[(m-n)^2(n^2-m) = 4m^2n\]

    So far, I havenīt found a single pair, but I cannot prove, that the set of solutions is empty.
    Perhaps, someone can help me to crack this nut?
    My suggestion is to take the whole thing a modulo 2 so you would have only a limited number of cases to check.

    -Dan

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    #3
    Ah well. It was a nice idea, but it doesn't actually help a whole lot. Good luck with it and if I get any other wild ideas I'll let you know.

    -Dan

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    #4
    Quote Originally Posted by lfdahl View Post
    Find the pairs of nonnegative integers, $(m,n)$, which obey the equality:

    \[(m-n)^2(n^2-m) = 4m^2n\]
    Not a solution but some elimination

    So far, I havenīt found a single pair, but I cannot prove, that the set of solutions is empty.
    Perhaps, someone can help me to crack this nut?
    first neither can be odd
    because if both are odd LHs is divisible by 8 being product of 3 even numbers and RHS is not

    if one is odd then LHS is odd and RHS even.

    thus both are even
    you can have this as a starting point

  5. MHB Oldtimer
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    #5
    Quote Originally Posted by lfdahl View Post
    Find the pairs of nonnegative integers, $(m,n)$, which obey the equality:

    \[(m-n)^2(n^2-m) = 4m^2n\]

    So far, I havenīt found a single pair, but I cannot prove, that the set of solutions is empty.
    Perhaps, someone can help me to crack this nut?
    Try for a solution with $m=2n$: $$(2n-n)^2(n^2-2n) = 16n^3,$$ $$n^3(n-2) = 16n^3,$$ $$ n-2 = 16,$$ $$ n=18.$$ That gives the solution $(m,n) = (36,18).$

    Edit. You can also try $m=3n$, which leads to the solution $(36,12).$
    Last edited by Opalg; November 8th, 2016 at 10:52.

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    #6
    Clever Opalg! That generalizes nicely.

    -Dan

  7. MHB Craftsman
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    #7 Thread Author
    Thankyou so much, Opalg, kaliprasad for your help to "crack the nut".
    Thankyou, topsquark for nice and helpful comments.

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