Thread: Find pairs of integers (m,n):

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

$(m-n)^2(n^2-m) = 4m^2n$

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

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

$(m-n)^2(n^2-m) = 4m^2n$

So far, I haven´t 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

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

4. Originally Posted by lfdahl
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 haven´t 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. Originally Posted by lfdahl
Find the pairs of nonnegative integers, $(m,n)$, which obey the equality:

$(m-n)^2(n^2-m) = 4m^2n$

So far, I haven´t 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 $(36,12).$

6. Clever Opalg! That generalizes nicely.

-Dan