Fermat proved that $x^4+y^4=z^4$ has no non-trivial solutions. I am sure that the diophantine equation below does have integer solutions if $a=b=c\neq \pm 1$ $$ax^4+by^4=cz^4$$ Now can one tell me how I can get all the possible integer solutions?
Parametric solutions $ax^4+by^4=cz^4$
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diophantine-equations
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0This isn't that easy as it looks. For the very first equation , there are no solutions, by FLT – 2017-02-27
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0There will be special triples for $(a,b,c)$ for which solutions exist Eg $(2,3,5)$ trivially ... more generally you will probably need to analyse on a case by case basis ... do you have any results about this yourself ? – 2017-02-27
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0Typo ... ? Fermat proved that $x^4+y^4=z^4$ **doesn't have** non-trivial solutions ... – 2017-02-27
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0As an indication of how hard this is, a question about the special case $(a,b,c)=(1,5,1)$ (http://math.stackexchange.com/questions/2025355/diophantine-equation-x45y4-z4#comment4163557_2025355) remains unanswered after more than 3 months. – 2017-02-27
2 Answers
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$ax^4+by^4=cz^4$ has parametric solution given below:
$$(a,b,c)=[(x^2),(2y^2-3x^2),(x^2+2y^2)]$$
Above has condition: $(z^2=y^2-x^2)$
Numerical solution for $(x,y,z)=(5,13,12)$ is given below,
$25(5)^4+263(13)^4=363(12)^4$
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Theorem about absence of non-trivial solutions of equation $$x^4 + y^4 = z^4$$ is a partial case of Fermat Last Theorem for exponent 4. It is a consequence of the single Fermat theorem with published author's prove.
The original question is too general for a clear answer.