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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?

  • This isn't that easy as it looks. For the very first equation , there are no solutions, by FLT – Aditya Narayan Sharma Feb 27 '17 at 19:42
  • There 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 ? – Donald Splutterwit Feb 27 '17 at 19:45
  • Typo ... ? Fermat proved that $x^4+y^4=z^4$ doesn't have non-trivial solutions ... – Adren Feb 27 '17 at 19:45
  • As 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. – Adam Bailey Feb 27 '17 at 22:31

2 Answers2

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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,

$25(5)^4+263(13)^4=363(12)^4$

M.Mass
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Sam
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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.

Yuri Negometyanov
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