# FLT by Freshman's dream

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### bearnol

Nov 14, 2013, 12:53:03 PM11/14/13
a^p+b^p+c^p=0
=> a^p+b^p+c^p==0 [mod p] for all p
=> a+b+c==0 [mod p] for all p
=> a+b+c=0

[only exception is (even) prime, p=2]
copyright JGW as of today 2013-11-14

### bearnol

Jul 22, 2016, 11:44:00 AM7/22/16
to Mersenneplustwo
FLT - the easy way - c JGW 2004-07-10
Inspired by http://www.bearnol.pwp.blueyonder.co.uk/Math/Cuboid.txt

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a^n+b^n+c^n=0

By symmetry, a solution a=b=c exists

but a, b, and c cannot all simultaneously be ==1 [mod 2] (by simple
congruences)

therefore a=b=c => a==b==c==0 [mod 2]

But if a==b==c==0 [mod 2] then there exists a smaller solution formed by
dividing out the factor of 2 from all of a,b,c

Hence, there is no smallest solution

Hence there is no integral solution

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It may amuse the reader to work out how the above reasoning doesn't
(quite) apply to the cases n=1, n=2