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31.4 Solving modular linear equations


Find all solutions to the equation $35x \equiv 10 \pmod{50}$.

$\{6, 16, 26, 36, 46\}$.


Prove that the equation $ax \equiv ay \pmod n$ implies $x \equiv y \pmod n$ whenever $\gcd(a, n) = 1$. Show that the condition $\gcd(a, n) = 1$ is necessary by supplying a counterexample with $\gcd(a, n) > 1$.

$$d = \gcd(ax, n) = \gcd(x, n)$$

Since $ax \cdot x' + n \cdot y' = d$,

we have

$$x \cdot (ax') + n \cdot y' = d.$$

$$ \begin{aligned} x_0 & = x'(ay / d), \\ x_0' & = (ax')(y / d) = x'(ay / d) = x_0. \end{aligned} $$


Consider the following change to line 3 of the procedure $\text{MODULAR-LINEAR-EQUATION-SOLVER}$:

 3  x0 = x'(b / d) mod (n / d)

Will this work? Explain why or why not.

Assume that $x_0 \ge n / d$, then the largest solution is $x_0 + (d - 1) \cdot (n / d) \ge d \cdot n / d \ge n$, which is impossible, therefore $x_0 < n / d$.

31.4-4 $\star$

Let $p$ be prime and $f(x) \equiv f_0 + f_1 x + \cdots + f_tx^t \pmod p$ be a polynomial of degree $t$, with coefficients $f_i$ drawn from $\mathbb Z_p$. We say that $a \in \mathbb Z_p$ is a zero of $f$ if $f(a) \equiv 0 \pmod p$. Prove that if $a$ is a zero of $f$, then $f(x) \equiv (x - a) g(x) \pmod p$ for some polynomial $g(x)$ of degree $t - 1$. Prove by induction on $t$ that if $p$ is prime, then a polynomial $f(x)$ of degree $t$ can have at most $t$ distinct zeros modulo $p$.