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

On Jul 14, 11:57 am, tommy1729 <tommy1...@xxxxxxxxx> wrote:

x^4+y^3=z^3

x,y,z > 1 , integers and relatively prime

tommy1729

y, z > 1 is an artificial distinction, because there's no
reason to exclude solutions with yz < 0. Not saying these
solutions definitely exist (although I'm sure they do);
but if so your condition will miss them.

Because your equation is inhomogenous, 'relatively prime'
is another artificial condition. Instead the condition to
exclude all but representative cases should be no integer
p such that p^4 divides (y, z) and p^3 divides x.

Your equation is equivalent to the following, in which we
can assume u is square-free:

z - y, z^2 + z.y + y^2 = u.v^2, u.w^2 [1.1]

u.v.w = x^2 [1.2]

Even if you're only interested in solutions with (y, z) = 1,
u can be any of -3, -1, 1, 3. I'll leave you to figure out
how the 3 pops up. (Hint - consider the expression I use
next for z^2 + z.y + y^2.)

Plugging the z - y expression into u.w^2 = (z - y)^2 + 3zy
gives:

3y.(y + u.v^2) = u.w^2 - u^2.v^4

<=> 3.(2y + u.v^2)^2 = u.(4.w^2 - u.v^4) [2]

If u can be any integer then since u is squarefree either
u or u/3 (if this is an integer) must divide 2y + u.v^2.

In the case, typically, for which u does not divide 3,
u divides 2y + u.v^2. So for some integer p we have:

2y = u.(p - v^2)

which in [2] gives:

3u.p^2 = 4.w^2 - u.v^4 [3]

But again since u is squarefree this implies that 2w = u.q
for some integer q, so that:

u.q^2 = v^4 + 3.p^2

and from [1.2] x = u.r for some integer r, giving:

v.q = 2.r^2

The latter implies one of the following for integers a, b, c
with 'a' squarefree:

v, q = 2a.b^2, a.c^2 _or_ a.b^2, 2a.c^2

r = a.b.c

Plugging the second (typically) into [3] gives the following,
from which we conclude a or (if 3 divides a) a/3 divides p :

4u.a^2.c^4 = a^4.b^8 + 3.p^2

Assuming (typically) that 3 does not divide a, so p = a.P
for some integer P, we obtain finally:

4u.c^4 = a^2.b^8 + 3.P^2

For any given u, if the following has non-trivial integer
solutions:

X^2 + 3.Y^2 = 4u.Z^2

then the general solution can be expressed in terms of one
or more sets of quadratics, homogenous in m, n, for X, Y, Z.

Each such set gives a solution to your original problem
if there is a solution to:

Z(m, n) = c^2

(Integer pair[s] a, b can always be found satisfying the
remaining condition X(m, n) = a.b^4)

The word 'typical' is used in the above to branch down one
of several (usually two) possibilities. As it suggests, the
analysis is similar for all cases but would be tedious to
detail in full.


Cheers

John R Ramsden

.

Relevant Pages

  • Re: x^4+y^3=z^3
    ... y, z> 1 is an artificial distinction, because ... reason to exclude solutions with yz < 0. ... p such that p^4 divides ... If u can be any integer then since u is squarefree ...
    (sci.math)
  • Re: x^4+y^3=z^3
    ... y, z> 1 is an artificial distinction, because ... reason to exclude solutions with yz < 0. ... p such that p^4 divides ... But again since u is squarefree this implies that 2w ...
    (sci.math)
  • Re: x^4+y^3=z^3
    ... y, z> 1 is an artificial distinction, because ... reason to exclude solutions with yz < 0. ... p such that p^4 divides ... But again since u is squarefree this implies that 2w ...
    (sci.math)