From - Sat Feb 7 20:50:04 1998
From: jpf@hydra.cfm.brown.edu (Jim P. Ferry)
Newsgroups: sci.math
Subject: Re: 3^x+5^y=7^z+1
Date: 6 Feb 1998 23:13:28 GMT
Organization: Brown University Center for Fluid Mechanics
|> The equation 3^x+5^y=7^z+1 has the solution (x=1,y=1,z=1).
|> Does another solution exist (x,y,z positive integer)?
|> If it doesn't, how to prove it?
Someone else posted this too (I think he said the problem is from an
old Czech olympiad), but I can't find his post. He was interested in
non-negative solutions, but it's trivial to show this adds only the
solution (0,0,0).
The solution below doesn't quite give all the details a proof would
require. Rather, it breaks up the problem into three steps, and
gives the result of each step, but without proof. This is laziness,
not pedagogy.
Let us consider the equation mod 3, 4, 5, 7, 9, 13 and 25. (I'm not
claiming to have the briefest possible solution!) We'll show there
are no further solutions.
Warmup: Considering the equation mod 3, 4 and 5 yields the following:
y = 1 mod 2, x = z = 1 mod 4.
Grist: Figurin' mod 7 and 9 and using the above yields:
x 1 5 12
x = 1 or y = 1 or 5 mod 6
z 5 9 12.
3^x mod 9 is the key: this separates out x = 1 as a special
case (which includes the known solution!): 5^y = 7^z - 2.
We now apply the same principle (i.e., mod p^2). It's easy to
see that the reduced equation has only the solution (1,1) by
considering it mod 25.
Okay, now we can hope to show the (mod 12, 6, 12) cases yield
no solution. The information we have will serve us best mod
some n such that phi(n) = 12.
Finale: Let's spell out this last consideration (mod 13):
0 1 2 3 4 5 6 7 8 9 10 11 (mod 12)
3^x 3 -4 (mod 13)
5^y 5 5 -5 -5 (mod 13)
7^z -2 -5 (mod 13)
So either 3 +/- 5 = -2 + 1 mod 13 or
-4 +/- 5 = -5 + 1 mod 13.
Neither works, so we're done.
-Jim Ferry