Question

\[x^y=y^x\] find interger pairs x and y ,hence show that the solution to \[x^2=2^x\] is \[x_1=2,x_2=4\]

Answer 1

obviously the immediate result is\[x=y=c ,\forall c \in \mathbb{R} \] but is the any other solution

Answer 2

i mean \[\forall c\in\mathbb{Z} \]

Answer 3

I think those are the only solutions, but I don't have a quick proof in mind. But I should note that you need to restrict to more than just \(c\in\mathbb{Z}\). You should restrict to just the positive integers, so \(c\in\mathbb{Z}^+\).

Answer 4

no wait, you don't have to restrict to that, I was just mixing the integers up with the real number. Whoops.

Answer 5

the fact that x=4 is a solution to the equation if y=2 should imply there is some solution
since
\[x=4\neq y=2\]
i tried to prove for the simple case when we have
\[\large x^2=2^x\]
set \[\color{blue}{x=2k}\]
\[\large 2^{2k}=(2k)^22^2=k^2\]
suppose k is of the form
\[k=2^a\] then
\[\large (2^2)^{2^a}=2^22^{2a}\]
\[\large (2^2)^{2^a-a-1}=1\]
\[2^a-a-1=0\]
since \[2^a\ge a+1\\ \forall a>1\in \mathbb{Z}\]
so
\[a=0,a=1\]

for \[\color{blue}{x=2k-1}\\2^{2k}=even=2(2k-1)^2=2(2k^2-k)+1=odd\]
contradition
so we dont have any odd solutions

Answer 6

equal sign misplaced in \[(2^2)^k=(2k)^2=2^2k^2\]

Answer 7

due to lack of tools transforming the equation by substitution\[x=\log_2t\]
\[t=(\log_2t)^2,t\ge0\] doesnt seem solvable atleast with elementary tools

Answer 8

\[t>0\]

Answer 9

It's similar to this, but I think this is a harder question.

http://en.wikipedia.org/wiki/Fermat%E2%80%93Catalan_conjecture