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I have modular polynomials that are of the form $$ x^n+C_1\equiv 0 \mod (C_2x+1). $$ and I am trying to find out information about the solutions for $x$ of is equation (if they exist). Note that I am not trying to determine if polynomial $C_2x+1 | x^n+C_1$, as that would be easy using polynomial division, rather if there is a way for determining the smallest value of $x$ of which this equation has a solution.

I understand that the statement provided is very general, but that is mostly because I am looking for resources to point me in the right direction of research such as

Is there a name for a polynomial mod a linear polynomial of the same variable? If not is there a name for an equation mod a polynomial?

Are there know methods or bounds on such polynomial solutions?

For example the polynomial equation $$ x^4+16 \equiv 0 \mod 128x + 1 $$ has minimal solution of $x=5$ apart from the trivial solution of $x=0$. Should additional structure be required, $C_1, C_2$ are guaranteed to be powers of $2$, however I'd like to see something more general

Bill Dubuque
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wjmccann
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1 Answers1

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Note $\bmod c_2 x + 1\!:\,\ \color{#c00}{x\equiv -1/c_2}\ $ so

$$0 \equiv \color{#c00}x^n+c_1\equiv \left[\color{#c00}{\frac{-1}{c_2\!}}\right]^n\!\!+c_1\equiv \frac{(-1)^n+c_1c_2^n}{c_2^n}\iff (-1)^n+c_1c_2^n\equiv 0\qquad$$

Or, we can eliminate use of fractions (or inverses) by scaling by $\,c_2^n\,$ at the start, i.e.

note $\gcd(c_2,c_2x+1)=1\,$ so scaling $\,0\equiv x^n+c_1\,$ by $\,c_2^n\,$ yields an equivalent congruence

$$\bmod \,\color{#c00}{c_2x+1}\!:\ 0\equiv (\color{#c00}{c_2 x})^n+c_1 c_2^n\equiv (\color{#c00}{-1})^n+c_1 c_2^n\qquad$$

So roots $\,x\,$ correspond to factors of $\,c_1 c_2^n+(-1)^n$ of form $\,c_2x+1\equiv 1\pmod{\!c_2},\,$ e.g. your

$$\bmod\, \color{#c00}{2^7 x+1}\!:\ 0\equiv (\color{#c00}{2^7 x})^4+ 2^4 (2^7)^4\equiv 1+2^{32}\qquad\quad\ $$

whose least factor $\,641$ is $\:\!\equiv 1\pmod{\!2^7},\,$ so $\,x = (641\!-\!1)/2^7 = 5,\,$ as you found.

Remark $ $ Regarding general methods, above is simply a special case of a generalization of the Factor or Remainder Theorem from monic factors $\,x\!-\!a\,$ to general linear factors $\,bx\!-\!a,\,$ where, as above, By Euclid, $\,(a,b)\!=\!1\Rightarrow (bn\!-\!a,b)=(a,b)\!=\!1,\,$ so we get an equivalent divisibility by scaling by $\,b^k\,,$ which keeps all arithmetic fraction-free (cf. here), i.e.

$$\begin{align}bn\!-\!a\mid f(n)&= f_k n^k + f_{k-1}n^{k-1} + \cdots + f_0\\[.3em] \iff bn\!-\!a\mid\, b^k f\ &= f_k (\color{#0a0}{bn})^k + f_{k-1} b (\color{#0a0}{bn})^{k-1}+\cdots + b^k f_0\\[.3em] &\equiv f_k\ \color{#c00}a^k\ \ +\ \ f_{k-1}\, b\,\color{#c00}a^{k-1} + \ \cdots\ \ + b^k f_0\!\!\pmod{\!\color{#0a0}{bn}\!-\!\color{#c00}a}\\[.3em] &= \,b^k f(a/b)\end{align}$$ $${\rm i.e.}\ \ \ \bbox[1px,border:1px solid #c00]{\bbox[6px,border:1px solid #c00]{bn\!-\!a\mid f(n) \!\iff\! \, bn\!-\!a\mid b^kf \left({\small \frac{a}b}\right)\in\Bbb Z\,}}\qquad\qquad\qquad\qquad\qquad\quad\!\!$$

Bill Dubuque
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