Can $a bmod 3$ be represented arithmetically without the mod or other integer-related functions?

Question

I've noticed that $a bmod b$ (with 'mod' in the operational sense) can also be represented using various tricks of formulation. For instance, that
$$a bmod b = a - bleftlfloorfrac{a}{b}rightrfloor .$$
There are several ways to do it by (ab)using various functions like ceiling, max, abs, and the like. However, I realized yesterday that
$$a bmod 2 = frac{1-(-1)^a}{2}.$$
I find this interesting as I consider basic exponentiation to be a purer operation in some sense than something like floor, perhaps in that it doesn't have a built-in sense of conditionality or knowledge of fractional parts. Furthermore, you can use substitution to reach arbitrarily high powers of $2$, as in
$$a bmod 4 = frac{1-(-1)^a}{2}+1-(-1)^{frac{a-frac{1-(-1)^a}{2}}{2}},$$
which amounts to right-shifting $a$ by one spot and repeating the process to get the second bit you need for the $bmod 4$. I realize that's not pretty, but I'm interested that it's possible. The motivation here is identifying situations where formulae may implicitly support a richness of computational complexity one wouldn't expect, which parity detection and manipulation goes a long way towards. Which leads me to my question:
Is there some way using exponentiation or other basic operations to find a comparable expression for $a bmod 3$, or ideally, $a bmod b$?

J. W. Tanner · Answer

Note that $sin(2pi n/3)=0, dfrac{sqrt3}2, $ or $-dfrac{sqrt3}2$, according as $nequiv0, 1, $ or $2bmod3$, respectively.
Furthermore, $f(x)=dfrac{xleft(x-dfrac{sqrt3}2right)2}{-dfrac{sqrt3}2left(dfrac{sqrt3}2-dfrac{sqrt3}2right)}+dfrac{xleft(x+dfrac{sqrt3}2right)1}{dfrac{sqrt3}2left(dfrac{sqrt3}2+dfrac{sqrt3}2right)}=dfrac{3x^2-dfrac{sqrt3}2x}{dfrac 32}=2x^2-dfrac x{sqrt3}$
has the property that $f(0)=0, fleft(dfrac{sqrt3}2right)=1, $ and $fleft(-dfrac{sqrt3}2right)=2$.
Therefore, $f(sin(2pi n/3))=2sin^2(2pi n/3)-dfrac {sin(2pi n/3)}{sqrt3}=n bmod 3.$

Greg Martin · Answer

Any function $f(n)$ on the integers that is periodic with period $m$ can be written in terms of powers of $zeta=e^{2pi i/m}$, a primitive $m$th root of unity:
$$
f(n) = sum_{k=0}^{m-1} hat f(k) zeta^{nk}, quadtext{where } 
hat f(k) = frac1m sum_{j=0}^{m-1} f(j) zeta^{-jk}.
$$
Notice that when $m=2$, we have $zeta=-1$ and this formula becomes
$$
f(n) = hat f(0) + hat f(1)(-1)^n, quadtext{where }
hat f(0) = frac12(f(0)+f(1)) text{ and } hat f(1) = frac12(f(0)-f(1));
$$
this is the formula you included when $f(n) = nmod 2$.
When $m=3$, we have $zeta=e^{2pi i/3} = frac12(-1+isqrt3)$ and $zeta^2=barzeta=frac12(-1-isqrt3)$, and
$$
f(n) = hat f(0) + hat f(1) zeta^n + hat f(2) barzeta^n.
$$
In the specific case $f(n)=nmod 3$, the coefficients are
begin{align*}
hat f(0) &= frac13(0+1+2) = 1, \
hat f(1) &= frac13(0+1barzeta+2zeta) = -frac12+frac i{2sqrt3}, \
hat f(2) &= frac13(0+1zeta+2barzeta) = -frac12-frac i{2sqrt3}.
end{align*}
Putting it all together,
$$
nmod3 = 1 + bigg({-}frac12+frac i{2sqrt3}bigg)bigg(frac{-1+isqrt3}2bigg)^n + bigg({-}frac12-frac i{2sqrt3}bigg)bigg(frac{-1-isqrt3}2bigg)^n.
$$

Can $a bmod 3$ be represented arithmetically without the mod or other integer-related functions?

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