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I am trying to solve the equation $(x-a)(x-b)\equiv 0 \pmod m$

And I am wondering if checking two equations $x-a\equiv k \pmod m$ and $x-b\equiv \frac{m}{k} \pmod m$ for all $k\mid m,$ would find all solutions $x$ under mod $m.$ (combined with the initial case where m divides $x-a$ or $m$ divides $x-b$)

Michael Hardy
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pblpbl
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  • Where you typed \;(mod \; m) I changed it to \pmod m, which achieves the same result more simply except that $\bmod$ is not italicized. Not that if more than one object is to be included in the parentheses after $\bmod,$ then you need $\big{$braces$\big}$, thus: \pmod{43} yields $\pmod{43},$ whereas \pmod43 yields $\pmod43.$ Note also the typographical difference between $k|m$ and $k\mid m,$ the latter coded as k\mid m. Likewise there is the symbol $k\nmid m,$ coded as k\nmid m. $\qquad$ – Michael Hardy Aug 06 '20 at 17:38

2 Answers2

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If $m$ is prime that all solutions of the quadratic equation will be found that way. But consider $$ (x- 2)(x-1)\equiv0\pmod6. $$ This has four distinct solutions: $0,1,2,4,5.$ When $x\equiv4$ then $(x-2)(x-1)\equiv 2\cdot3\equiv0$ and when $x\equiv5$ then $(x-2)(x-1) \equiv 12\equiv0.$

The reason this can happen with a composite number like $6$ is you can multiply two non-zero numbers and get $0,$ thus $2\cdot3\equiv 0.$

The usual quadratic formula works when the modulus is an odd prime number (as opposed to the one even prime number, $2$), except that dividing by $2a$ has to be construed as multiplying by the multiplicative inverse of $2a$ modulo the prime number in question, and instead of saying $b^2-4ac$ is positive, you have to say $b^2-4ac$ has a square root. Half of all non-zero (congruence classes of) numbers will have square roots.

Michael Hardy
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I believe the answer is no, but the difference between the algorithm you proposed and what would work is a little more subtle than described in Michael Hardy's answer.

It seems that you have tried to account for non-zero zero divisors (in other words numbers that multiply to zero) by including the possibility that $x-a\equiv k\pmod{m}$ and $x-b\equiv m/k\pmod{m}$ for some factor $k$ of $m$. This does produce solutions like $x\equiv 4$. However, this does not cover all cases, because zero divisors in $\mathbb{Z}/m\mathbb{Z}$ aren't all factors of $m$, they can also be factors of a multiple of $m$. In fact, all numbers that are not coprime to $m$ are zero divisors. For example as Michael Hardy points out, $4$ is not a factor of $6$, but $4\times 3\equiv 12\equiv 0\pmod{6}$, and note that $\gcd(4,6) = 2$.

Therefore to fix your algorithm, you should consider all zero divisor pairs $p$ and $q$, i.e. $pq\equiv 0\pmod{6}$, and check $x-a\equiv p\pmod{m}$ and $x-b\equiv q\pmod{m}$.

Elliot Yu
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