 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  (x−y)^ n = xy (Posted on 2017-01-07) For what triplets (x, y, n) of positive integers is the equation (x−y)^ n = xy ?

 No Solution Yet Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 2 of 2 | First n=1.  Then x-y = x*y.  This can be rearranged as (x+1)*(y-1) = -1.  There are then two integer solutions (x,y) = (0,0) and (-2,2), but these both fail the postive requirement.

Now n=2.  Then (x-y)^2 = x*y.  Adding 4xy to each side and rearrange to get (x+y)^2 = 5*x*y.  Then x*y and 5*x*y are both perfect squares.  But that can only happen when xy=0.  So the only solution in this case is (x,y)=(0,0) and that also fails the positive requirement.

Now for n>2, the 'large' values of n.  Let g be the GCD of x and y.  Then let a and b be the numbers so that x=a*g and y=b*g.  a, b, and a-b will all be coprime to each other

Substitute the expressions in the equation to yield g^(n-2) * (a-b)^n = a*b.  a-b must be a factor of the right side, but the right side is coprime to a-b.  This can only happen when a-b=1.

Substitute a=b+1. Then the equation becomes g^(n-2) = b*(b+1).  If n=3 then g = b*(b+1), which forms the parameterization (x,y) = (b*(b+1)^2, b^2*(b+1)).  This describes the set Charlie found in his numeric search.

If g>3 then g^(n-2) is a perfect power.  Since b and b+1 are coprime, they would need to both be perfect powers of n-2 for the equation to be solvable in integers.  However there are no positive consecutive perfect powers so no positive solutions exist for n>3.  (Aside from the trivial x=y=0 which fails the positive requirement.)

In summary, all positive integer triplets (x,y,n) which satisfy the equation are (b*(b+1)^2, b^2*(b+1), 3) for all positive integer b.

 Posted by Brian Smith on 2017-01-07 20:44:53 Please log in:

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