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

Mathematics Asked on January 5, 2022

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, $$a bmod b$$?

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, $$f(sin(2pi n/3))=2sin^2(2pi n/3)-dfrac {sin(2pi n/3)}{sqrt3}=n bmod 3.$$

Answered by J. W. Tanner on January 5, 2022

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, $$nmod3 = 1 + bigg({-}frac12+frac i{2sqrt3}bigg)bigg(frac{-1+isqrt3}2bigg)^n + bigg({-}frac12-frac i{2sqrt3}bigg)bigg(frac{-1-isqrt3}2bigg)^n.$$

Answered by Greg Martin on January 5, 2022

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