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$$x^3 +y^4 =z^5$$
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.

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    $\begingroup$ When you say "solve", do you mean find solutions that are real numbers, integer numbers or rational numbers? $\endgroup$
    – guest
    Nov 4, 2014 at 6:46

2 Answers 2

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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.

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

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    $\begingroup$ I need a Generalized way . $\endgroup$
    – Fazla Rabbi Mashrur
    Nov 4, 2014 at 6:38
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    $\begingroup$ Not an exact answer . $\endgroup$
    – Fazla Rabbi Mashrur
    Nov 4, 2014 at 6:39
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    $\begingroup$ sorry don't know $\endgroup$
    – thunder
    Nov 4, 2014 at 6:39

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