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Marquez's Equation (Posted on 2005-03-02) Difficulty: 3 of 5
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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Solution 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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