My first reaction is x=y. However, a more unified answer is harder to find. this is developing as I type... so maybe I'll make a mistake or get stuck (forgive me if I do Please Register or Log in to view the hidden image! ) let y=x+a x^y = y^x <=> x^(x+a) = (x+a)^x <=> x^x + x^a = x^x + x*ax^(x-1) + ...+x*xa^(x-1)+a^x (Pascal's sequence) <=> x^a = x*ax^(x-1)+ 2(x-1)*a^2*x^(x-2)+...+2(x-1)*a(x-2)*x^2+x*xa^(x-1)+a^x <=> some cleaning up to do... x^a = ax^x + ... Hmmm I need a paper now. I'll get back to this later.
does this mean you have the answer now? or do I have to continue? (or were you merely sarcastic?) Please Register or Log in to view the hidden image!
No........ I don't have the answer... yet. But you gave me a starting point. continue, of course! Let's see who gets first to the answer. James R: I don't know the answer. Trying to find it, though. I took Marlijns' example and tried doing what he did but it seemed too much to "clean up" so I assumed y=ax and I got x^(ax)=(xa)^x || \/ (x^x)^a=x^x*a^x || \/ (x^x)^(a-1)=a^x And then I did ln[(x^x)^(a-1)]=ln(a^x) || || (a-1)*x*lnx = x*lna || \/ lnx=(lna)/(a-1) and then e^(lnx)=e^[(lna)/a-1)] || || x = (a-1)st root of a = a^[1/(a-1)] try it on a calculator. It actually works. Now, this is true for real numbers, but I need to find all a's for which the expression a^[1/(a-1)] = Z is true. Merlijn, try to finish what you started. Maybe that will give the answer.
BloodSuckingGerbile, Ok, I will try to work on it; it's a lot of cleaning up though. Maybe there is an easier way. However, I seriously doubt that y=ax makes it any easier. bye
I think I have the answer. (a-1)*x*lnx = x*lna (before I divided lna by a-1) Gives the answer a=1, or y=x*1=x, which is true. Then when I divided lna by (a-1) I got lnx=(lna)/(a-1) and then x=a^[1/(a-1)] So there is only one integer a for which a^[1/(a-1)] is an integer and that is a=2, which gives x=2 and y=a*x=2*2=4. :bugeye: Please Register or Log in to view the hidden image! Just one more thing: -2 and -4 are also true... %@#$%!!!!!! Please Register or Log in to view the hidden image! I guess I'll have to do that from the beginning...
well you know that with your y=ax there is a solution a=1 (for all x) and a=2 (at least for x=2) negative solutions for a are higly improbable (hmmm x=1, a=-1 gives 1/1 = -1 ? eeeh obviously not! Please Register or Log in to view the hidden image!) Forget the negative solutions.
well only if both x and y are negative... but not only one of the two... I haven't had time yet to have another look at the problem... be patient pleace. (I am sure there are others who can solve the problem)
Cool problem, gerbil Please Register or Log in to view the hidden image! I think your approach works well, but you have to take care. As such: y = ax, x^y = y^x x^(ax) = (ax)^x (x^a)^x = (xa)^x But this is where we have to pay attention. If x is odd, then we can just raise both sides to power 1/x to obtain: x^a = xa, x odd If x is even then we have two possibilities: x^a = xa, and x^a = -xa. In either case, if a < 0 then |1/x^|a|| = |xa| => 1/|a| = |x^(|a|+1)|. Since x and a are integers, 1/|a| must be an integer which can only be if |a|=1, but a is even. Therefore a>0. Note that this applies regardless of whether x is odd or even (which means y and x must have the same sign.) In the second case, let's consider the possible values of a: a=2: x^2 = -2x => x = -2, y = -4 a=4: x^4 = -4x => x^3 = -4 => -1 < x < -2 a=6: x^6 = -6x => x^5 = -6 => -1 < x < -2 ... as a approaches infinity, x approaches -1 but is never again an integer. Which means the only solution for this case is x=-2,y=-4. Now let's return to the other case: x^a = xa, a>0 So let's try some values of a: a=1: x = x => y = x a=2: x^2 = 2x => x = 2, y = 4 a=3: x^3 = 3x => x^2 = 3 => 1 < x < 2 a=4: x^4 = 4x => x^3 = 4 => 1 < x < 2 ... In the limit, as a becomes infinite x approaches 1 but is never again an integer. Which means that all values of a greater than 2 are excluded. So your solution is: a=1: y=x, x any integer (positive or negative) a=2: x=2,y=4 and x=-2,y=-4 That's it as far as integer solutions go, unless I missed something...
Overdoze, thanks for fully answering my question. Not being careful is my biggest problem. Please Register or Log in to view the hidden image!
I was just playing around with Mathematica to see what it can do. It came up with the solution: -x * ProductLog[ -Log[x]/x ] ________________________ Log[x] This is an interesting function: for 0<=x<=e it evaluates to x, at e it has its maximum, after which it decreases and has limit 1. So the only Integer solutions this function gives at y=2 are x=2 and x=4. Other integer solutions are x=y=0 and x=y=1. GRAPH Why Mathematica misses the obvious x=y solution is a mystery to me. Maybe I don't know the program well enough.
hmm... That sounds interesting. I've never heard of Mathematica. Some kind of software? Tell me more, please.
Mathematica is the program from Wolfram Research that can do symbolic mathematics. See http://www.wolfram.com Greetings, Han.
