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>