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Exponential Diophantine Equation

Date: 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
Associated Topics:
College Number Theory
High School Number Theory

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