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I know it is true that, if for example, $a \equiv b \equiv 1 \bmod 2$, then $$a^3 + ab^2 + b^3 \equiv 1^3 + 1 \cdot 1^2 + 1^3 \equiv 1 \bmod 2.$$ (I hope this notation is fine, but if not, someone please correct me.)

I'm trying to understand exactly why this is the case. If I worked purely with equivalence classes, I can say that $[a] = [b] = [1]$, so $$ [a^3] + [ab^3] + [b^3] = [a]^3 + [a][b]^3 + [b]^3 = [1]^3 + [1][1]^3 + [1]^3 = [1] $$ by the rules for adding and multiplying equivalence classes. Alternatively, I could use the following rules iteratively: $$(a \equiv b \bmod n) \wedge (c \equiv d \bmod n) \implies a + c \equiv b + d \bmod n$$ $$(a \equiv b \bmod n) \wedge (c \equiv d \bmod n) \implies ac \equiv bd \bmod n.$$ So the proof would look like this. We have $a \equiv 1 \bmod 2$. Applying the second rule twice, I get $a^3 \equiv 1 \bmod 2$. Similarly, $b^3 \equiv 1 \bmod 2$. By the first rule, $a^3 + b^3 \equiv 0 \bmod 2$. By the second rule, $b^2 \equiv 1 \bmod 2$ so $ab^2 \equiv 1 \bmod 2$. By the first rule again, $a^3 + ab^2 + b^3 \equiv 1 \bmod 2$.

My question is: is there another way to understand the shorthand I wrote down in the second line above, or are these the ways to understand it?

Bill Dubuque
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    Why would you need another way? Isn't that simple and clear as it is? But okay, another way to understand it $\equiv 1 \pmod 2$ means "odd" and $\equiv 0 \pmod 2$ means "even". So if $a$ and $b$ are both odd then $a^3 +ab^2 + b^3$ is odd. That's all it is saying..... But actually, if we weren't familiar with odd/even $\pmod 2$ should be as easy as doing arithmetic. Knowing that if $a\equiv k \pmod n$ and $b\equiv j \pmod n$ then $a\circ b \equiv k\circ j \pmod n$ for all basic arithmetic functions (not division and not exponents) should be intuitive and utterly simple to understand. – fleablood Jul 07 '22 at 03:19
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    We can break it down a bit further by noting that any product of odd numbers is odd, and any sum of an odd number of odd numbers is odd. – Gerry Myerson Jul 07 '22 at 04:23
  • See the Polynomial Congruence Rule in the linked dupe. By inductive extension to $n$ variables that yields $,\forall i!:,x_i\equiv \bar x_i\Rightarrow f(x_1,\ldots,x_n) \equiv f(\bar x_1,\ldots,\bar x_n),,$ for any polynomial $f$ with integer coef's. – Bill Dubuque Jul 07 '22 at 06:58

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