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Post Posted: Mar 03, 2008, 8:51 am • # 1 


Let p be a prime number. Solve in \mathbb{N}_0\times\mathbb{N}_0 the equation x^3+y^3-3xy=p-1.

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Post Posted: Mar 03, 2008, 10:58 am • # 2 


x^3 + y^3 - 3xy + 1 = (x + y + 1)(x^2 + y^2 + 1 - x - y - xy) = p
x = y = 1 doesn't give any solution so x + y > 0. hence x^2 + y^2 + 1 - x - y - xy = 1
if x and y are bigger than 1; \frac {x^2 + y^2}{2} > = x + y} and \frac {x^2 + y^2}{2} > = xy so x^2 + y^2 > = xy + x + y and in the equality case x = y = 2 gives p = 5
if x or y is 1 say x = 1. y^2 - 2y = 0 so y = 2 or y = 0. checking yields only y = 0 is a solution.
so the solutions are (2,2,5), (0,1,2), (1,0,2)
 
 
Post Posted: Mar 07, 2008, 7:14 pm • # 3 


this reminds me of the problem:
for which n does the equation n=x^3+y^3+z^3-3xyz have a solution. :)

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Post Posted: Apr 05, 2008, 7:29 am • # 4 


And what is the solution of problem that Albanian Eagle wrote ? Is it well-known ? :maybe:

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