Diophantine Equation with Three UnknownsDate: 09/28/2009 at 22:48:52 From: Hue Subject: Diophantine Equation with 3 unknowns Find all triples of positive integers (x,y,z) such that xyz + xy + yz + zx + x + y + z = 243 Here's what I've tried: xyz + xy + yz + zx + x + y + z = 243 = 3^5 (xyz + xy + yz + zx + x + y + z) --------------------------------- = 1 243 1 1 1 1 1 1 1 1 ---- + ------ + ------ + ------ + ------ + ------ + ------ = ------ 243 243z 243x 243y 243yz 243xz 243xy xyz 1 1 1 1 1 1 1 1 ---- - ------ - ------ - ------ - ------ - ------ - ------ = ------ xyz 243z 243x 243y 243yz 243xz 243xy 243 If x, y, and z are each at least 2, then the L.H.S will be smaller than the R.H.S which is less than 1/243. So that means that at least one of the variables must be smaller than 2, right? If x = 1, 1 1 1 1 1 1 1 1 ---- - ------ - ------ - ------ - ------ - ------ - ------ = ------ yz 243z 243 243y 243yz 243z 243y 243 1 1 1 1 1 1 2 ---- - ------ - ------ - ------ - ------ - ------ = ------ yz 243z 243y 243yz 243z 243y 243 1 1 2 2 2 ---- (1 - ------) - ------ - ------ = ------ yz 243 243y 243z 243 From here I have no idea how to continue. Thanks. Date: 09/29/2009 at 02:21:05 From: Doctor Greenie Subject: Re: Diophantine Equation with 3 unknowns Hi, Hue -- Thanks for sending an interesting problem. I didn't like the looks of the path you started down towards solving the problem, so I went a completely different direction. xyz + xy + xz + yz + x + y + z = 243 x(yz + y + z) + (yz + y + z) = 243 - x (x + 1)(yz + y + z) = 243 - x yz + y + z = (243 - x)/(x + 1) The left-hand side must be an integer, so (x+1) must divide (243-x): (x+1) | (243 - x) (x+1) | (243 - x) + (x + 1) (x+1) | 244 This gives us only the following possible values for x+1: 2, 4, 61, 122 which means x can only be 1, 3, 60, or 121 We can then separate the problem into four cases with these values of x. For each case, we encounter another problem that is solved by the same technique as above. (1) x = 1... yz + y + z = (243 - 1)/(1 + 1) = 242/2 = 121 y(z + 1) + z = 121 y(z + 1) = 121 - z y = (121 - z)/(z + 1) (z+1) | (121 - z) (z+1) | (121 - z) + (z + 1) (z+1) | 122 z + 1 = 2 or 61 z = 1; y = (121-1)/(1+1) = 120/2 = 60 or z = 60; y = (121-60)/(60+1) = 61/61 = 1 So we have the solution (1,1,60) (and permutations of that solution, if we are looking for ordered triplets). (2) x = 3... yz + y + z = 240/4 = 60 y(z + 1) + z = 60 y(z + 1) = 60 - z y = (60 - z)/(z + 1) (z+1) | (60 - z) (z+1) | (60 - z) + (z + 1) (z+1) | 61 z + 1 = 61 z = 60 y = 0/61 = 0 The triple (0,3,60) gives us a solution to the equation, but it is not a solution in positive integers. So we still have only the one solution. I will let you finish the cases for x = 60 and x = 121. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ Date: 09/29/2009 at 05:05:56 From: Hue Subject: Thank you (Diophantine Equation with 3 unknowns) Hi Doctor Greenie, Thanks for your guiding!!! - Hue |
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