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 Marquez's Equation (Posted on 2005-03-02)
Find all postive integer solutions to the equation x^y-y^x=x+y.

 See The Solution Submitted by owl Rating: 4.1429 (7 votes)

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 re(2): sorry to say...- there it goes !!! | Comment 8 of 9 |
(In reply to re: sorry to say... by owl)

`I repeat that I didn't achieve this proof.`
`I could never do that !!!! `
`And since owl authorized to do, I'm posting it.`
`Perhaps, just perhaps, there is a more easy`
`way, that I didn't find.`
`The equation is<o:p></o:p>`
`<o:p></o:p>`
`x^y - y^x = x + y.<o:p></o:p>`
`<o:p></o:p>`
`Now let's check negative numbers first.<o:p></o:p>`
`<o:p></o:p>`
`If x <= 0 and y <= 0, then we multiply both `
`sides of the equation by<o:p></o:p>`
`(x+y)x^(-y)y^(-x) and factor<o:p></o:p>`
`<o:p></o:p>`
`( (x+y)x^(-y) - 1 )( (x+y)y^(-x) + 1 ) = -1,`
`where each factor is an integer. `
`Then we either have<o:p></o:p>`
`<o:p></o:p>`
`(x+y)x^(-y) = (x+y)y^(-x) = 0,<o:p></o:p>`
`<o:p></o:p>`
`which gives x = -y (and then the hypothesis `
`requires x = y = 0), or<o:p></o:p>`
`<o:p></o:p>`
`(x+y)x^(-y) = 2<o:p></o:p>`
`(x+y)y^(-x) = -2.<o:p></o:p>`
`<o:p></o:p>`
`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<o:p></o:p>`
`combinations, and none work.<o:p></o:p>`
`So, x and y can't BOTH be negative.<o:p></o:p>`
`<o:p> </o:p>`
`If x >= 0, then x^y = y^x + x + y is an `
`integer, so x = 1 or y >= 0.<o:p></o:p>`
`And x = 1 implies 1 - y = 1 + y, thus y = 0.`
`Indeed, (1, 0) is a solution. `
`In any case, y >= 0.<o:p></o:p>`
`If y >= 0, then y^x = x^y - x - y `
`is an integer, so y = 1 or x >= 0.<o:p></o:p>`
`And y = 1 implies x - 1 = x + 1,`
`which has no solutions.`
`Thus x >= 0.<o:p></o:p>`
`<o:p></o:p>`
`So we conclude that, apart from (0, 0),`
`we have x > 0 and y > 0.<o:p></o:p>`
`<o:p></o:p>`
`Now we look at the positive solutions,`
`and we ask which is bigger: x or y?<o:p></o:p>`
`<o:p> </o:p>`
`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<o:p></o:p>`
`<o:p></o:p>`
`0 < x + y = x^y - y^x <= 0.<o:p></o:p>`
`<o:p></o:p>`
`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.<o:p></o:p>`
`We already saw that y = 1 implies`
`x - 1 = x + 1, which has no solutions.<o:p></o:p>`
`If y = 0 and x > 0, then 1 = x + y,`
`thus x = 1, and we have the solution (1, 0).<o:p></o:p>`
`Therefore, (0, 0) and (1, 0) are the only `
`solutions with x >= y.<o:p></o:p>`
`<o:p> </o:p>`
`Now we conclude that 0 < x < y.<o:p></o:p>`
`We already checked x = 1, which gives y = 0.`
`So 1 < x < y.<o:p></o:p>`
`We want to show that x^y - y^x is usually `
`much bigger than x + y.<o:p></o:p>`
`We might have trouble when x = 2. `
`So we'll save that for later.<o:p></o:p>`
`Fix x >= 3, and let<o:p></o:p>`
`<o:p></o:p>`
`g(y) = x^y - y^x - x - y, so that<o:p></o:p>`
`<o:p></o:p>`
`g'(y) = (x^y)(log y) - (x)(y^(x-1)) - 1.<o:p></o:p>`
`<o:p></o:p>`
`I want to show that g'(y) > 0 when y > x.`
`Recall that y > x >= 3 implies<o:p></o:p>`
`<o:p></o:p>`
`(log x)/x > (log y)/y<o:p></o:p>`
`y log x > x log y.<o:p></o:p>`
`<o:p> </o:p>`
`Also, y > x implies y >= x + 1,`
`and therefore log y >= log (x + 1),<o:p></o:p>`
`<o:p></o:p>`
`and y > x >= 3 implies that log log y > 0.<o:p></o:p>`
`<o:p></o:p>`
`Adding these together gives<o:p></o:p>`
`<o:p></o:p>`
`y log x + log y + log log y > `
`x log y + log (x + 1).<o:p></o:p>`
`<o:p> </o:p>`
`Anti-logging gives<o:p></o:p>`
`<o:p></o:p>`
`x^y * y * log y > y^x * (x + 1).<o:p></o:p>`
`<o:p></o:p>`
`Then we divide by y and get<o:p></o:p>`
`<o:p></o:p>`
`(x^y)(log y) > y^(x-1) * (x + 1)<o:p></o:p>`
`             = (x)(y^(x-1)) + y^(x-1)<o:p></o:p>`
`             > (x)(y^(x-1)) + 1<o:p></o:p>`
`<o:p></o:p>`
`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. <o:p></o:p>`
`<o:p></o:p>`
`Therefore, if y > x >= 3, then<o:p></o:p>`
`<o:p></o:p>`
`x^y - y^x - x - y = g(y) >=`
`g(x+1) = x^(x+1) - (x+1)^x - x - (x+1).<o:p></o:p>`
`<o:p> </o:p>`
`Now I want to show that this is bigger `
`than zero.  So let<o:p></o:p>`
`<o:p></o:p>`
`h(x) = x^(x+1) - (x+1)^x - x - (x+1).<o:p></o:p>`
`= e^((x+1)(log x)) - e^(x log(x+1)) - 2x - 1<o:p></o:p>`
`<o:p> </o:p>`
`So that its derivative is<o:p></o:p>`
`<o:p></o:p>`
`h'(x) = (x^(x+1))(log x + 1 + 1/x) - `
`      - ((x+1)^x)(log(x+1) + 1 - 1/(x+1)) - 2<o:p></o:p>`
`<o:p> </o:p>`
` = (x^(x+1))(log x) - ((x+1)^x)(log(x+1))<o:p></o:p>`
` + x^(x+1) + x^x - (x+1)^x + (x+1)^(x-1) - 2<o:p></o:p>`
`<o:p> </o:p>`
` > (x^(x+1) - (x+1)^x)(log x) +<o:p></o:p>`
` (x^(x+1) - (x+1)^x) + x^x + (x+1)^(x-1) - 2<o:p></o:p>`
`<o:p> </o:p>`
`which is positive since x >= 3 implies<o:p></o:p>`
`<o:p></o:p>`
`x^(x+1) - (x+1)^x > 0,log x > 0, x^x > 2.<o:p></o:p>`
`<o:p> </o:p>`
`Therefore,h(x) is always increasing for x >= 3.`
`And<o:p></o:p>`
`<o:p></o:p>`
`h(3) = 3^4 - 4^3 - 3 - 4 = `
`= 81 - 64 - 3 - 4 = 10 > 0,<o:p></o:p>`
`<o:p> </o:p>`
`which means that h(x) > 0 for all x >= 3.`
`Therefore,`
`we have no solutions with y > x >= 3.<o:p></o:p>`
`<o:p></o:p>`
`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.<o:p></o:p>`
`<o:p></o:p>`
`We want to find its roots. `
`As before, take the derivative<o:p></o:p>`
`<o:p></o:p>`
`g'(y) = (2^y)(log 2) - 2y - 1.<o:p></o:p>`
`<o:p></o:p>`
`A graphing calculator shows that this function`
`is negative for y values from zero to `
`something between 3 and 4, and <o:p></o:p>`
`<o:p></o:p>`
`then it becomes positive. `
`So first let's show that if y >= 4,`
`then g'(y) > 0. `
`By taking another derivative, we get<o:p></o:p>`
`<o:p></o:p>`
`g"(y) = (2^y)(log 2)^2 - 2<o:p></o:p>`
`<o:p></o:p>`
`and then this is positive since y >= 4 `
`(in fact, y > 3) implies<o:p></o:p>`
`<o:p></o:p>`
`2^y > 2^3 = 8<o:p></o:p>`
`(2^y)(log 2)^2 - 2 > `
`> (2^y)(1/2)^2 - 2 > (8)(1/4) - 2 = 0.<o:p></o:p>`
`<o:p> </o:p>`
`Therefore, g'(y) is increasing when y > 3.`
`So when y >= 4, then<o:p></o:p>`
`<o:p></o:p>`
`g'(y) >= g'(4) = (2^4)(log 2) - 2*4 - 1<o:p></o:p>`
`<o:p></o:p>`
`and this is positive as long as<o:p></o:p>`
`<o:p></o:p>`
`log 2 > 9/16,<o:p></o:p>`
`<o:p></o:p>`
`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<o:p></o:p>`
`that g(y) > 0 when y > 5. `
`So now we check y values less than 5 (that<o:p></o:p>`
`is, y=1, 2, 3, and 4 when x = 2),`
`and we find that there are no other<o:p></o:p>`
`solutions.<o:p></o:p>`

 Posted by pcbouhid on 2005-06-03 14:43:41

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