Re: x^4+y^3=z^3

The reference for coprime x, y, z is Darmon and Granville (Bull LMS,
1995). If you allow common factors, as you say, it is easy to cook up
infinitely many examples, at least provided one of the exponents is
coprime to the others. For example, if we want to find solutions to

x^p+y^q=z^r

where, say, r is coprime to pq, then, writing s for the least common
multiple of p and q and setting

c = a^p+b^q

for our favourite a and b, we have solutions

x=a c^(ks/p), y= b c^(ks/q) and z = c^((1+ks)/r)

for each positive integer k for which

ks = -1 mod r.

For whatever reason, people tend to regard these solutions as
"uninteresting". I'm not sure that's fair (and these are certainly not
all the solutions!).

All the best,

Gaew

On Jul 15, 2:54 am, OwlHoot <ravensd...@xxxxxxxxxxxxxx> wrote:
On Jul 15, 10:32 am, gaew <benn...@xxxxxxxxxxx> wrote:





On Jul 14, 2:06 pm, tommy1729 <tommy1...@xxxxxxxxx> wrote:

No (prime)integers can justify!

proof ?

tommy1729

In general, for fixed positive integers p, q and r with 1/p+1/q+1/r <
1, the equation

x^p+y^q=z^r

has at most finitely many solutions in nonzero, coprime integers x, y
and z. In the case you are considering, it has been proved (By Bruin)
to have none.

All the best,

Gaew

That may well be so for _coprime_ integers x, y, z in Tommy's
equation (and the more general class of equations you quoted -
a reference would be handy).

But it is easy to prove that sum{i=1..n}{x_i^p_i} = 0 (n >= 2)
must have non-zero integer solutions if any one of p_i is prime
relative to all the others.

Cheers

John R Ramsden


.

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