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Re: Fermat revised ?
Posted:
Aug 17, 2007 10:45 PM


> more clearly from x^2 + y^3 = z^4 > lets go to y^3 = z^4  x^2 = (z^2 +x)(z^2 x) > now we can conclude, that number y should be > built from two factors: let y = a*b > and a^3 = z^2 + x > b^3 = z^2  x > and then 2z^2 = a^3 + b^3................(*) > 2x = a^3  b^3 > and where only the eq(*) will determine some > solution: we can notice for a=4 b=2; > 2z^2 = 64 + 8 = 2*36 and z=6;x=28;y=8 > however even 28^2 + 8^3 = 6^4 > so it equals (2^4)(7^2) + 2^6 = (2^4)(3^4) > and after extraction of 2^4 will be only: > 7^2 + 2^2 = 3^4 > Also for to achieve primitive solutions it is to > add as in Your developments: condition a and b as > odd numbers... > so far I used just to calculate set: > x=433; y=13*11=143; z=42 > Thank You very much for more bigger samples. > You mentioned also, that there is no certain method for > to achieve such solutions ? > Anyway to the author of this topic "stephane" > should be said that there are no primitive solutions > from n=3: > any x^3 + y^4 equals z^5 for x;y;z of gcd=1 and etc. Also so called "Fermat revised" > x^n + y^(n+1) = z^(n+2) conjecture for n=>3 > is just a part of well known Beal's conjecture... > With the Best Regards > Roman B. Binder > POBox 3692 Jerusalem 91035 Israel > RoBin >
Thanks for your answer. Btw, I wasn't aware of the Beal's conjecture. I will check it out certainly.
First, I wasn't really looking for solution which are relatively prime ( I just read in the web that if n=2 the equation has solution only if x, y and z are relatively prime ... This is not the case as I could see from the first answer...).In fact, I was simply looking for solutions. Since, I have able to find solutions for n=3 X Y Z 256 64 32 268912 19208 2744 209952 11664 1944 839808 23328 3888
I found out also that I could find solutions in negative X Y Z 128 32 16 33614 2401 343 69984 3888 648 559872 15552 2592
I was unable to find pattern for these solutions.
SF 



