Can anyone help me explain why if $x$ is an integer, then $(x^3+1)\bmod 3 = (x+1)^3 \bmod 3$? I know there are 3 cases. $x=0\bmod3,\ x=1\bmod3,$ and $x=2\bmod3$ totally new to this form of mathematics, could anyone help me setup the proof?
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If you're allowed to use it, by Fermat's Little Theorem $n^3\equiv n\pmod3$. Using this gives
$$x^3+1\equiv(x+1)^3\equiv x+1\pmod 3$$

Mike
- 13,318
pmod
) suggests that a congruence was intended. At least in the first equation where binary mod (\bmod
) should be used. I do speak (loudly) against ever using binary mod myself, so this is kinda kettle calling pot, but..? Ok, the original post wasn't entirely clear either :-) – Jyrki Lahtonen Mar 02 '15 at 07:14mod
as a remainder operation. However, your syntax is not consistent thoughout the post, and at some points it looks like you might intend a congruence instead. Please study this Wikipedia page that explains the differences, and explain in your own words how you see it. The reason I'm asking is that your interpretation has an impact on how to the equations/congruences should best be typeset. It is also IMO beneficial to you to be aware of the differences! – Jyrki Lahtonen Mar 02 '15 at 08:15mod
are closely related, but the syntax is quite different. As is the semantics: binary mod is an arithmetic operation on two numbers that gives a third number as an output. a congruence could be viewed as having three arguments (the last one usually fixed throughout the question) and outputs a boolean value (true/false) in a sense. – Jyrki Lahtonen Mar 02 '15 at 08:18