4
$\begingroup$

Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.

Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...

$\endgroup$
4
$\begingroup$

This is a case of the generalized Fermat equation $$ x^p+y^q=z^r. $$ For $(p,q,r)=(3,4,2)$ we have $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1$, which is the spherical case. Here we have infinitely many integer solutions for this triple. The solutions are given by a finite set of polynomial parametrisations of the equation, see the following paper:

F. Beukers, The diophantine equation $Ax^p + By^q = Cz^r$, Duke Math.J. 91(1998), 61-88.

Further Reference: The generalized Fermat equation.

$\endgroup$
3
$\begingroup$

Here is one simple parameterization. We have,

$$x^4 +(y^2-1)^3 = (y^3+3y)^2$$

given the Pell equation $x^2-3y^2 =1$.

$\endgroup$
  • $\begingroup$ @AlexD $r=2$ in this question. $\endgroup$ – Dietrich Burde Dec 29 '18 at 23:34
  • $\begingroup$ Thanks, @DietrichBurde. Let me try again: A solution to the problem here is a tuple of 3 numbers. But the parameterization here only relates $x$ and $y$. What about $z$? $\endgroup$ – Alex D Dec 30 '18 at 11:21
  • $\begingroup$ @AlexD We have $X^4+Y^3=Z^2$ and $X$,$Y$,$Z$ depend on $x$ and $y$ only. So no $z$ in here, only two parameters $x,y$ here for an equation in three variables $X,Y,Z$. $\endgroup$ – Dietrich Burde Dec 30 '18 at 11:47
  • $\begingroup$ Well, @DietrichBurde, you can find out more, but how can you find a general formula (solve this equation in general form), expressed in arbitrary numbers (for example) m, n, k? For example, find the top three numbers (6, 5, 29)? $\endgroup$ – Yan Dashkow Dec 30 '18 at 11:55
  • $\begingroup$ @YanDashkow We find them using one of the parametrizations, see Beukers article. $\endgroup$ – Dietrich Burde Dec 30 '18 at 13:31
0
$\begingroup$

Above equation shown below has parameterization:

$x^3+y^4=z^2$

The below parameterization has no restriction such as the

Pell equation condition demonstrated by Tito Piezas.

$x=(p)^2(-q)^3$

$y=(p)(q)^2(k-1)$

$z=(p)^2(q)^4(2k-3)$

where, $p=(k-2)$ and $q=(k^2-2)$

For $k=3$ we get : $(-343)^3+(98)^4=(7203)^2$

$\endgroup$
  • $\begingroup$ I have a question: through which k you can find three solutions (x, y, z) — (6,5,29)? $\endgroup$ – Yan Dashkow Dec 30 '18 at 10:55
0
$\begingroup$

"OP" asked for parametric solution for $(x,y,z)=(6,5,29)$ in $x^3+y^4=z^2$

Solution is:

$x=3p^3(8k^2-40k+50)$

$y=p^2(20k^2-104k+135)$

$z=p^4(2k-5)^2(116k^2-540k+621)$

Where, $p=(4k^2-27)$

For $k=(13/5)$, we get after removing common factors:

$6^3+5^4=29^2$

$\endgroup$
  • $\begingroup$ Wait, why can't k be whole? $\endgroup$ – Yan Dashkow Dec 30 '18 at 19:30
0
$\begingroup$

"OP" enquired about integer coefficent's for the parametric

solution for the equation $(x^2+y^4=z^2)$. "OP" just needs

to substitute $k=(m/n)$ in the parametrization & the resulting

solution after removing common factors is given below.

$x=6(u^3)(v^2)$

$y=(u^2)(v)(10m-27n)$

$z=(u^4)(v^2)(116m^2-540mn+621n^2)$

And $u=(4m^2-27n^2)$ & $v=(2m-5n)$

For $(m,n)=(13,5)$ we get:

$6^3+5^4=29^2$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.