Exponential Diophantine EquationDate: 06/24/2005 at 02:54:04 From: Itzik Subject: a^a*b^b=c^c a,b,c are integers grater then 1. I'm looking for 3 integers a,b,c > 1 such that a^a * b^b = c^c. I think the numbers should be like this: (x^p*y^q)^[x^p*y^q] * (x^m*Y^n)^[x^m*y^n] = (x^t*y^c)[x^t*y^c] I found one answer: a = 12^6 b = 6^8 c = 2^11*3^7 I'm wondering if there are others? Date: 06/24/2005 at 22:20:13 From: Doctor Vogler Subject: Re: a^a*b^b=c^c a,b,c are integers grater then 1. Hi Itzik, Thanks for writing to Dr. Math. I checked and confirmed that there are no solutions with both a and b less than 1000 using the math program GNU Pari, http://pari.math.u-bordeaux.fr/ and the program m=2;n=2;count=1000 for(i=1, count, m=m+1; if(m>n,m=2;n=n+1); L=n*log(n)+m*log(m); k=0; u=n+m; l=L/log(u); while(floor(u+0.000000000001) > floor(l-0.00000000001), k=k+1; u=L/log(l); l=L/log(u)); print([n,m,k,l,u])) But I think that there are more answers beyond the one you found. Let p be a prime number, and let p^r be the largest power of p that divides a. Similarly let p^s and p^t be the largest powers of p that divide b and c. Then we have ar + bs = ct. Furthermore, this equation has to hold for every prime number. Of course, r=s=t is a solution regardless of a,b,c, so we only need to consider those primes that are factors of one of those three numbers. In your case, for p = 2, we get r = 12, s = 8, c = 11, and 12(12^6) + 8(6^8) = 11(2^11*3^7), which is true, though this is hard to tell until you divide by the gcd(a, b, c) = 2^8 * 3^6. After some more work on this problem, I found six more solutions, and there was a pattern. I found the pattern and came up with infinitely many solutions. It is possible that I have found all of the solutions with a and b bigger than 1, but I am not sure. If you set g = 2^(2^(n+1)*(2^n - 1 - n)) * (2^n - 1)^(2(2^n - 1)), that is, n+1 n n (2 (2 - 1 - n)) n (2(2 - 1)) g = 2 * (2 - 1) , (this is the GCD) and then set a = g * (2^n - 1)^2 b = g * 2^(2n) c = g * 2^(n+1) * (2^n - 1), then you have a solution for every positive integer n. When n=1, you just get a trivial (a=1, b=c) solution. When n=2, you get the solution you described. When n>2, you get different solutions, which get really big really fast. But you can substitute these formulas into your equation and check that it satisfies the equation for every n. The way I found these was by first singling out the GCD, g. If you set g = GCD(a, b), then you can fairly easily verify that g divides c as well (after you establish that c < a+b). Then you can divide a, b, and c by g and get a' = a/g b' = b/g c' = c/g a'^a' * b'^b' * g^(a'+b'-c') = c'^c', which is the same as your original equation except for the power of g. Then I noticed that your solution has a' + b' - c' = 1. That means that, as long as a'^a' divides c'^c', and b'^b' also divides c'^c' (remember that they are relatively prime, so that means their product also divides c'^c'), then you can set g = c'^c'/(a'^a' * b'^b'). Now if we change the order of a' and b' (possibly), then we can assume that a' < b'. Then you can show that b' must divide (a' - 1)^2. So I had a computer look through all integers a' up to 100,000, have b' check all factors of (a' - 1)^2 bigger than a', and then I checked if a'^a' and b'^b' both divided (a'+b'-1)^(a'+b'-1). It found six solutions in addition to the one you mentioned. The pattern in the b's was pretty obvious, and the pattern in the a's was not terribly hard to find. I also had a computer look through a' + b' - c' = m for integers m up to 100, checking a' up to 10,000, and found only a handful of solutions where a'^a' and b'^b' divided c'^c', and a' and b' are relatively prime (all of them with b' a power of 4), and none of them gave g^m values that were m'th powers. It caused me to conjecture that a and b relatively prime with a^a dividing c^c and b^b dividing c^c implies that b is a power of 4. But I quickly realized that this was absurd, since you can pick any a and b and then let c be the huge number a^a*b^b. In any case, we have infinitely many solutions, and they might be all of them, but they might not. There may well be solutions other than the ones I found. I am only claiming that the ones I described are infinitely many solutions. And I couldn't find any other solutions. 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: 07/02/2005 at 15:06:10 From: Itzik Subject: Thank you (a^a*b^b=c^c a,b,c are integers grater then 1.) Hi, Doctor Vogler - Thank you for your effort. Your answer is very nice. Regards, Itzik |
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