Timeline for Disproving Euler proposition by brute force in C

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Oct 31, 2016 at 15:55	comment	added	David Hammen		@MartinBonner -- Taking modulo 3 says that either none of a, b, and c can be multiples of 3, or that exactly two of a, b, and c must be multiples of 3. Modulo 5 is a bit more helpful: Exactly two of a, b, and c must be multiples of 5.
Oct 27, 2016 at 14:57	history	edited	mascoj	CC BY-SA 3.0	correction to logic
Oct 26, 2016 at 10:52	comment	added	Martin Bonner supports Monica		So with @DanUznanski simplification, if a is odd, both b and c must be even, if a is even, b can be either, and c must be the opposite parity to b.. That means the search space is cut in just over half. Can we do similar tricks with mod 3 calculations?
Oct 26, 2016 at 1:42	comment	added	Dan Uznanski		Actually it gets even better than this: all even numbers, when taken the to the fourth power, are divisible by 16, and all odd numbers, when taken to the fourth power, have a remainder of 1 when divided by 16. So it must be -- since having all of a, b, c, and d even means there's a smaller solution -- that d must be odd and exactly one of a, b, and c must be odd as well, because that's the only way to get the same remainder mod 16 on both sides.
Oct 26, 2016 at 0:19	review	First posts
Oct 26, 2016 at 3:53
Oct 26, 2016 at 0:17	history	answered	mascoj	CC BY-SA 3.0