Solving the Diophantine Equation x^y - y^x = x + yDate: 04/30/2005 at 08:48:50 From: Paulo Subject: math It was presented to me, to be solved in integers, the following equation: x^y - y^x = x + y. The two obvious solutions, just by inspection, are (x,y)=(1,0) and (2,5). My feeling is that there are no more solutions, but how can that be proved? Or, if there are more solutions, how do we find them? Thank you for your attention. Date: 04/30/2005 at 15:55:55 From: Doctor Vogler Subject: Re: math Hi Paulo, Thanks for writing to Dr. Math. This is called a Diophantine Equation because you are looking for integer solutions (although some reserve the word for polynomial equations only). It is not an easy one to solve, although it is solvable. It is similar to the related problem which you can find in our archives Solving the Equation x^y = y^x http://mathforum.org/library/drmath/view/66166.html Your equation is x^y - y^x = x + y. Now let's check negative numbers first. If x <= 0 and y <= 0, then we multiply both sides of the equation by (x+y)*x^(-y)*y^(-x) and factor ( (x+y)x^(-y) - 1 )( (x+y)y^(-x) + 1 ) = -1, where each factor is an integer. Then we either have (x+y)x^(-y) = (x+y)y^(-x) = 0, which gives x = -y (and then the hypothesis requires x = y = 0), or (x+y)x^(-y) = 2 (x+y)y^(-x) = -2. But this requires x^(-y) divisible by 2, hence x divisible by 2. Thus x = -1 or x = -2. Similarly, y = -1 or y = -2. We try all four combinations, and none work. So x and y can't BOTH be negative. If x >= 0, then x^y = y^x + x + y is an integer, so x = 1 or y >= 0. And x = 1 implies 1 - y = 1 + y, thus y = 0. Indeed, (1, 0) is a solution. In any case, y >= 0. If y >= 0, then y^x = x^y - x - y is an integer, so y = 1 or x >= 0. And y = 1 implies x - 1 = x + 1, which has no solutions. Thus x >= 0. So we conclude that, apart from (0, 0), we have x > 0 and y > 0. Now we look at the positive solutions, and we ask which is bigger: x or y? If x >= y >= 3, then (log x)/x <= (log y)/y, since f(x) = (log x)/x is a decreasing function for x > e, and therefore y log x <= x log y, and x^y <= y^x, so 0 < x + y = x^y - y^x <= 0. If y = 2 and x > 0, then we still get (log x)/x <= (log y)/y except when x = 3, but (3, 2) is not a solution. We already saw that y = 1 implies x - 1 = x + 1, which has no solutions. If y = 0 and x > 0, then 1 = x + y, thus x = 1, and we have the solution (1, 0). Therefore, (0, 0) and (1, 0) are the only solutions with x >= y. Now we conclude that 0 < x < y. We already checked x = 1, which gives y = 0. So 1 < x < y. We want to show that x^y - y^x is usually much bigger than x + y. We might have trouble when x = 2. So we'll save that for later. Fix x >= 3, and let g(y) = x^y - y^x - x - y, so that g'(y) = (x^y)(log y) - (x)(y^(x-1)) - 1. I want to show that g'(y) > 0 when y > x. Recall that y > x >= 3 implies (log x)/x > (log y)/y y log x > x log y. Also, y > x implies y >= x + 1, and therefore log y >= log (x + 1), and y > x >= 3 implies that log log y > 0. Adding these together gives y log x + log y + log log y > x log y + log (x + 1). Anti-logging gives x^y * y * log y > y^x * (x + 1). Then we divide by y and get (x^y)(log y) > y^(x-1) * (x + 1) = (x)(y^(x-1)) + y^(x-1) > (x)(y^(x-1)) + 1 which is equivalent to g'(y) > 0. Therefore, when x >= 3, the smallest value of g(y) with y > x occurs when y = x + 1. Therefore, if y > x >= 3, then x^y - y^x - x - y = g(y) >= g(x+1) = x^(x+1) - (x+1)^x - x - (x+1). Now I want to show that this is bigger than zero. So let h(x) = x^(x+1) - (x+1)^x - x - (x+1). = e^((x+1)(log x)) - e^(x log(x+1)) - 2x - 1 So that its derivative is h'(x) = (x^(x+1))(log x + 1 + 1/x) - ((x+1)^x)(log(x+1) + 1 - 1/(x+1)) - 2 = (x^(x+1))(log x) - ((x+1)^x)(log(x+1)) + x^(x+1) + x^x - (x+1)^x + (x+1)^(x-1) - 2 > (x^(x+1) - (x+1)^x)(log x) + (x^(x+1) - (x+1)^x) + x^x + (x+1)^(x-1) - 2 which is positive since x >= 3 implies x^(x+1) - (x+1)^x > 0, log x > 0, x^x > 2. Therefore, h(x) is always increasing for x >= 3. And h(3) = 3^4 - 4^3 - 3 - 4 = 81 - 64 - 3 - 4 = 10 > 0, which means that h(x) > 0 for all x >= 3. Therefore, we have no solutions with y > x >= 3. Finally, we have to check x = 2. We already know that there is one solution with x = 2. Now consider g(y) = 2^y - y^2 - y - 2. We want to find its roots. As before, take the derivative g'(y) = (2^y)(log 2) - 2y - 1. A graphing calculator shows that this function is negative for y values from zero to something between 3 and 4, and then it becomes positive. So first let's show that if y >= 4, then g'(y) > 0. By taking another derivative, we get g"(y) = (2^y)(log 2)^2 - 2 and then this is positive since y >= 4 (in fact, y > 3) implies 2^y > 2^3 = 8 (2^y)(log 2)^2 - 2 > (2^y)(1/2)^2 - 2 > (8)(1/4) - 2 = 0. Therefore, g'(y) is increasing when y > 3. So when y >= 4, then g'(y) >= g'(4) = (2^4)(log 2) - 2*4 - 1 and this is positive as long as log 2 > 9/16, which it is (9/16 = 0.5625, and log 2 = 0.6931...). Therefore, g(y) is increasing for y >= 4. As you already know, g(5) = 0, which means that g(y) > 0 when y > 5. So now we check y values less than 5 (that is, y=1, 2, 3, and 4 when x = 2), and we find that there are no other solutions. And there you have it! If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/