Re: A diofantine equation

gerry@xxxxxxxxxxxxxx wrote:
Mate wrote:
I seems that the equation 7^x + 2 = y^2
does not have integer solutions for x,y>1.
Can you suggest a method to prove this?

There are standard methods for tackling this kind of problem.
Unfortunately I have to look
them up whenever I want to use them, and I am currently 10,000 miles
from my references.
One place to look would be Mordell's book on diophantine equations.
Another place would
be several papers written by Leo J Alex over 20 years ago on
exponential diophantine
equations.

Here's something that might lead to a solution, though I give no
guarantees.
Write 7^x = y^2 - 2 = (y + s) (y - s), where I use s as an abbreviation
for square root of 2.
Z[s] is a unique factorization domain. 7 factors in Z[s] as 7 = (3 + s)
(3 - s). So
(3 + s)^x (3 - s)^s = (y + s) (y - s). Since Z[s] is a UFD, this almost
means
(3 + s)^x = y + s; I say almost, because there may be a unit involved
(also there are
some simple relative primality conditions which you have to check). The
fundamental
unit in Z[s] is 1 + s, so we also have the possibility that (3 + s)^x =
(1 + s) (y + s).
And maybe there are some other possibilities involving associates of 1
+ s.

Let's look at (3 + s)^x = y + s. Expand the left side by the binomial
theorem and compare
coefficients of s; 1 = x 3^(x - 1) + (x choose 3) 2 3^(x - 3) + ... +
2^r, where x = 2 r + 1
(others have already pointed out that we may assume x is odd). Now
comes the hard
part - trying to derive a contradiction from this equation (together
with the assumption
r > 0). Sorry, that's all I can do without my books.



Thank you for the suggestion.
The case (3 + s)^x = y + s can be easily discarded because obviously
the coefficient of s in LHS is >1.
The case (3 + s)^x = (1 + s) (y + s) can be also be rejected because
it leads to [(3 + s)^x - (3 - s)^x] / (2s) = y+1 and the LHS is odd
for x odd and y must be odd also.

Unfortunately Z[s] has many units and the case
(3 + s)^x = (1 + s)^k (y + s) must be also eliminated
for _each_ k in Z. It seems to me that this is almost
as difficult as the original question.

Mate.

.

Relevant Pages

  • Re: A diofantine equation
    ... does not have integer solutions for x,y>1. ... There are standard methods for tackling this kind of problem. ... One place to look would be Mordell's book on diophantine equations. ... (others have already pointed out that we may assume x is odd). ...
    (sci.math)
  • Re: A diofantine equation
    ... does not have integer solutions for x,y>1. ... There are standard methods for tackling this kind of problem. ... One place to look would be Mordell's book on diophantine equations. ...
    (sci.math)
  • Re: The Structure of Primitive Pythagorean Progressions
    ... > rer wrote: ... >> are two possible scenarios: ... >> others three must be even and e will be odd, if there are to be integer ... >> to be integer solutions ...
    (sci.math)
  • Re: --- --- --- Reference related to a number theory questio
    ... david wrote: ... x, y, z are relatively prime integers each> 5, prime p> ... But it can have integer solutions if M is odd. ...
    (sci.math)
  • Re: A Diophantine Equation
    ... consider the integer solutions of the equation ... both odd (because otherwise they must both be even and ... from the formula for primitive pythagorean triples. ...
    (sci.math)