Find All Pairs of Integer Solutions to a^(2b) = b^aDate: 02/18/2005 at 18:58:31 From: Jason Subject: Powers of Powers Investigate pairs of positive integers (a,b) that satisfy the equation a^(2b) = b^a. I do not really know where to start, to be honest! I have tried lots of different values for a and b but have yet to find one that satisfies the equation. Date: 02/19/2005 at 09:50:18 From: Doctor Vogler Subject: Re: Powers of Powers Hi Jason, Thanks for writing to Dr. Math. Are you a college student? If so, I would recommend going to the university library and finding the January 2004 edition of the math journal "American Mathematical Monthly," where they had an article that discussed how to solve y^x = x^(my) for positive integers m. It gave the results for m = 1, 2, 3 in particular. If not, write back and I'll try to give you some hints at how to solve it. But first, do you know how to solve the equation WITHOUT the 2? - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 02/21/2005 at 18:54:51 From: Jason Subject: Powers of Powers Hello Doctor Vogler, I would just like to say a big thank you for your help, it is much appreciated. I have managed to find the math journal, so feel I can understand how to solve the equation without the 2 as asked but am having difficulty getting to grips with the generalization equation. Would it be possible to explain it in a simpler way? A few things also puzzle me: For the equation x^y = y^x it only had 1 integer solution of (2,4). Does the equation y^x = x^(2y) also have limited solutions? The generalization equation in the journal seems to concentrate on finding positive rational numbers but I just want the ones where (x,y) are positive integers that satisfy the equation so I am getting confused with some of the solutions. For instance, x^y = y^(mx), take logs, substitute y = x^r, we get x = (mr)^(1/(r-1)) and y = (mr)^(r/(r-1)), page(15). However, substitute m = 2 and substitute positive integers for r. I miss some values of (x,y) that are integers such as (9,27). Now if I use the equations x = (2+(2/n))^n, y = (2+(2/n))^(1+n), page(16) using positive values of n=1,2,3,... I find (9,27) but miss out (2,16)! Is there a solution that just finds positive integers for x and y? Hope you can follow this! I am getting a little confused so any help you could give me would be great. Thank you again. Jason Date: 02/22/2005 at 22:13:54 From: Doctor Vogler Subject: Re: Powers of Powers Hi Jason, You're right. The techniques can be slightly different for integers only. I was asked this a little while ago without the 2, and I gave the following answer. Look it over, and see if you can apply the techniques for the integer solutions in the case with the 2. Tell me what you get, and if you get stuck, then I'll look it over and see if I can take it any further. The question was: Find all positive integers a and b such that a^b = b^a. My reply was: Thanks for writing to Dr. Math. The short answer to your question is that all solutions in positive integers are a = b and (2, 4) or (4, 2). The long answer, of course, is a proof of the same. Before going there, however, I would like to point out that your idea is a terrific idea for finding all *real* solutions. Allow me to elaborate. Since the function f(x) = (log x)/x increases from x=0 to x=e and then decreases from x=e to infinity, if 0 < a < b <= e or if e <= b < a, then we have f(a) < f(b) and therefore (log a)/a < (log b)/b b log a < a log b a^b < b^a. But if we have a < e < b, then that is not enough information to tell whether a^b or b^a is bigger. And for each a < e there is exactly one b > e where f(a) = f(b) and therefore a^b = b^a, and for smaller values of b (but still bigger than e) you will have a^b < b^a while for larger values of b you will have a^b > b^a. But this idea of using continuous functions becomes less useful when dealing with integer or rational solutions, since the Intermediate Value Theorem doesn't apply. So we need to try a different idea. Try this: Substitute t = b/a into your equation. If a and b are integers, or if a and b are rational numbers, then t will be rational. Substituting a*t for b, we get t a a (a ) = (at) . Now, taking the (positive real) a'th root of both sides of the equation, we get t a = at and therefore t-1 a = t and 1/(t-1) a = t . In fact, we can get a parametrized form for all rational solutions from this equation. Let n = 1/(t-1), so that a = (1 + 1/n)^n b = (1 + 1/n)^(n+1). These are rational solutions for all (nonzero) integers n. It takes a little more work to show that these (along with a=b) are *all* rational solutions. And that work begins with proving the lemma: If r/s and c/d are rational numbers in lowest terms, and (r/s)^(c/d) is also rational, then r and s are both d'th powers of integers. But you wanted integer solutions. And I'm done with my tangents. The first step is to prove that t is an integer. Then all that's left is a simple induction argument. Of course, t is not an integer if b=2 and a=4. But if we permit ourselves to switch a with b (note that this doesn't change the equation) so that a <= b then we can prove that t will be an integer. You see, if a <= b, then a b a divides a . But a^b = b^a, which means that a a a divides b , and this is enough to imply that a divides b (think about the previous statement in terms of the prime factorizations of a and b), and therefore t=b/a is an integer. And if a and b are positive, then t is also positive. Now t=1 gives the solutions with a=b. And if t=2, then t-1 a = t gives a = 2 (and therefore b = 4). And a=1 quickly implies b=1. But if a >= 2, then t-1 t-1 a >= 2 , and we can prove by induction that t-1 2 > t for all integers t >= 3. This is easily verified for t=3. Now suppose that t-2 2 > t-1 and then we have t-1 t-2 2 = 2 * 2 > 2(t-1) = t + (t-2) > t. Therefore, there are no solutions with t > 2, and so we have found all of the integer solutions, namely (k,k), (2,4), and (4,2). If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 02/23/2005 at 07:29:21 From: Jason Subject: Powers of Powers So if x^y = y^(mx), now if m>1 (in this case 2) x,y > 0. Now if x,y does not =1, then by taking logs we get log(x)/x = mlog(y)/y substitute for y = x^r where r and x do not equal 1. For a start, I am confused at how you get to this (below) so cannot find an equation in terms of r to start with. Then x = (mr)^(1/(1-r)) and y = (mr)^(r/(1-r)) But for integer solutions, we must prove that r is an integer. Obviously r cannot be an integer if x and y are integer solutions. x < y, x and y cannot be equal. The equation is not satisfied if y < x If r = 2y/x = p/q is an integer, if x, y are positive, then r is also positive. I don't think this is correct at all. r will never equal 1 as x < y are never the same. I don't really know what I doing, I can nearly follow through your proof by induction but how does this prove what the integer solutions are, it just proves that there are no integer solutions where t > 2, doesn't it? Could you show me how to work out the integer solutions for x x^y = y^(2x) as it not coursework, we are merely looking at different investigations and modelling exercises within the course. Thank you again for all your help, sorry to be such a pain. Many thanks! Jason Date: 02/23/2005 at 11:51:18 From: Doctor Vogler Subject: Re: Powers of Powers Hi Jason, All right, let me sit down and think about this for a bit. We want to avoid logs if we need integer solutions (or rational ones, for that matter) since the log of an integer is not an integer, not rational, and generally indistinguishable from the log of any other real number. Recall that we set t = y/x and then, when we wanted integer solutions, we tried to show that t could be made an integer (by possibly switching x and y). So it all comes down to finding the t values. In the case with m=1, we tried to show that x had to divide y, which meant that t had to be an integer. Can we do something similar here? One step we CANNOT do is to conditionally switch x and y, because now that WILL change the equation. But let's think about this anyway. We have x^y = y^(mx) and we want to know if x has to divide y, or perhaps vice-versa. Let's consider a few different cases. Case 1: mx <= y In this case, clearly x^(mx) divides x^y which equals y^(mx), and then we can take (mx)'th roots and therefore x divides y. Case 2: y <= mx In this case, y^y divides y^(mx) which equals x^y, and then we can take y'th roots and therefore y divides x. Hmm! So now we *almost* have that t is an integer. In fact, we have that either y/x is an integer, or x/y is. That means we need to check both possibilities. So let's suppose that t = y/x is an integer. Then y = tx and we can substitute this into our equation and get x^(tx) = (tx)^(mx). Now we take the positive real x'th root of each side of the equation and get x^t = (tx)^m or x^t = t^m * x^m. This is essentially the same thing we did when m was 1. Let's continue what we did, which was to put all of the x's on the left side. So we divide both sides of the equation by x^m: x^(t-m) = t^m. Now what we did in the m=1 case is show that when x and t are large (larger than 2 or 3), then this couldn't possibly happen. You can do the same thing when m=2, or any other m, for that matter. But you'll have to use bigger x and t values when m is bigger. In other words, first you check the x=1 case, which always gives you t^m = 1 and therefore t=1. Well, x=1 and y=1 is certainly a solution. If x>1 and t<m, then the left side will not be an integer, while the right side is, so there are no solutions there. If t=m, then again you get t^m = 1 and therefore t=1, contradicting t=m (unless m=1). So then you have to look at t>m. You suppose x>1 (so x>=2) and first check t=m+1: x^(1) = (m+1)^m and you find that you always have a solution x = (m+1)^m y = (m+1)^(m+1). In particular, for m=2 this is (9, 27). Then you try to prove that this won't happen for any larger values of t. So we assume that x >= 2 and therefore x^(t-m) >= 2^(t-m) and we try to prove by induction on t that 2^(t-m) > t^m. But actually this isn't true for just any m. But it will be true when t is large enough. In fact, depending on m, you'll have to check the first several values of t (from m+2 up until the above inequality is satisfied, which happens at t=m+7=9 when m=2 for example). Then the inductive step is the same, since if t > 1 + 1/(2^(1/m) - 1) then 2(t-1)^m > t^m and so the inductive hypothesis 2^(t-m-1) > (t-1)^m implies that 2^(t-m) = 2 * 2^(t-m-1) > 2(t-1)^m > t^m which does it, once we check the smaller values of t. For each smaller value of t, we have to check if x = t^(m/(t-m)) is an integer. That is, if t^m is a perfect (t-m)'th power. For m=2, you need to do this for t=4 to t=8, and you get 4^(2/2) = 4 5^(2/3) 6^(2/4) 7^(2/5) 8^(2/6) = 2 and the other three values are not integers. So that means that you have two other solutions, namely t=4 and t=8, which have x = 4, y = 16 and x = 2, y = 16. And these, together with x = 9, y = 27, will be all solutions in positive integers for the equation x^y = y^(2x). If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 02/27/2005 at 07:37:55 From: Jason Subject: Thank you (Powers of Powers) I would just like to say a huge thank you to Dr. Vogler for answering all of my questions, it is much appreciated. Keep up the good work. Best wishes, Jason |
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