3
$\begingroup$

$$x^3 +y^4 =z^5$$
How can I solve this equation.I only know trial and error method, but it's not a generalized way.
Please tell me a generic way to solve this type of equation.

$\endgroup$
  • 1
    $\begingroup$ When you say "solve", do you mean find solutions that are real numbers, integer numbers or rational numbers? $\endgroup$ – guest Nov 4 '14 at 6:46
4
$\begingroup$

The equation $x^3 + y^4 = z^5$ has infinitely many solutions in positive integers. An infinite family of solutions is generated by $$x = a(a^3 + b^4)^{8k},\qquad y=b(a^3 + b^4)^{6k},\qquad z=(a^3 + b^4)^{5k}.$$ There are probably other solutions; I doubt that an exhaustive list of solutions is known.

Beal's conjecture would imply that the equation has no relatively prime integer solutions. But this conjecture remains unproved, and there is a $1 million prize for a proof or counterexample.

$\endgroup$
-1
$\begingroup$

$$2^{24} +2^{24}=2^{25}$$
$$(2^{8})^{3} + (2^{6})^{4}=(2^{5})^{5}$$
So,$$ x=256\ ,\ y=64 \ ,\ z=32$$

$\endgroup$
  • 1
    $\begingroup$ I need a Generalized way . $\endgroup$ – Fazla Rabbi Mashrur Nov 4 '14 at 6:38
  • 1
    $\begingroup$ Not an exact answer . $\endgroup$ – Fazla Rabbi Mashrur Nov 4 '14 at 6:39
  • 1
    $\begingroup$ sorry don't know $\endgroup$ – thunder Nov 4 '14 at 6:39

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.