imqwerty
  • imqwerty
another ques..
Mathematics
katieb
  • katieb
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imqwerty
  • imqwerty
l,m,n are positive real numbers such that\[l^3+m^3=n^3\]prove that\[l^2+m^2-n^2>6(n-l)(n-m)\]
MrNood
  • MrNood
I am not sure about this - but I was under the impression that there were NO solutions for that equation (isn't this Fermat's last theorem problem?)
ganeshie8
  • ganeshie8
Fermat's last theorem was about "integer" solutions

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mathmath333
  • mathmath333
\(\color{white}{bookmark}\)
AbdullahM
  • AbdullahM
Didn't @kainui solve this yesterday?
imqwerty
  • imqwerty
thas ws not the proper solution :)
MrNood
  • MrNood
@ganeshie8 ah yes - sorry my mistake
imqwerty
  • imqwerty
ok :) heres the solution-
thadyoung
  • thadyoung
1 Attachment
imqwerty
  • imqwerty
( ͡° ͜ʖ ͡°)=found
ParthKohli
  • ParthKohli
That's a crazy solution. You can't really ever normally think of this.
imqwerty
  • imqwerty
solution obtained by normal thinking - u can write that inequality like -\[l^2+m^2-n^2-6(n-l)(n-m)\]\[7n^2-6(l+m)n-(l^2+m^2-6lm)<0\] x=7c^2 y=-6(l+m)n z=-(l^2 +m^2 -6lm) so u have to prove x+y+z<0 u can also observe that x ,y, z are not all equal x>0 , y<0 use the identity- \[x^3+y^3+z^3-3xyz=\frac{ 1 }{ 2 }(x+y+z)[(x-y)^2 +(y-z)^2+(z-x)^2]\] but its enough to prove that x^2 + y^3+z^3 -3xyz<0 substitute x, y,z :/ i didn't tell this cause i hate the tedious calculations it involves after some simplification this will reduce to-\[-l^2m^2(129l^2+129m^2-254lm)<0\] notice that this thing does is not having n cause i substituted n^3=m^3+l^3 while simplifying and clearly \[129l^2 +129m^2 -254lm=129(l-m)^2+4lm>0 \] hence the results follow ...
ParthKohli
  • ParthKohli
haha normal thinking
imqwerty
  • imqwerty
:D

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