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For any $p$ prime, $b \in \{1, ..., p-1\}$ and $a \in \mathbb{N}$,
why is there always a $x \in \mathbb{N_0}$ such that $b \mid px + a$?

Bill Dubuque
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David Vielhuber
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  • It is not that $px+a$ is always divisible by $b$. For most $x$ it won't be... but we will always be able to find one such that it is. – JMoravitz Mar 09 '22 at 01:40
  • Thanks, I clarified the title. – David Vielhuber Mar 09 '22 at 01:42
  • Are you familiar with bezout's identity? How about the euclidean division algorithm? Quotient remainder theorem? – JMoravitz Mar 09 '22 at 01:44
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    Suppose $px+a\equiv 0\pmod{b}$. Then $px\equiv -a\pmod{b}$. Now... to solve for $x$ if this were just a normal equality of real numbers, we'd say "divide both sides by $p$ to get $x=-\frac{a}{p}$", but its not a normal equation... it is a modular equivalence. Are you aware of the fact that we can still "divide" though in modular equivalencies under the right circumstances (by way of multiplicative inverses)? Does $p$ have a multiplicative inverse modulo $b$? It should... because $p$ is prime. How do we know what it is? Bezouts identity and the extended euclidean division algorithm. – JMoravitz Mar 09 '22 at 01:57
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    By the proof in the linked dupe: $,px\equiv -a\pmod{!b},$ has an integer root $x\iff \gcd(p,b)\mid a,,$ which is true here since the gcd $=1.,$ The root is $,x\equiv -a/p \equiv -ap^{-1}\pmod{!b},,$ which can be computed by various methods linked there. – Bill Dubuque Mar 09 '22 at 08:51

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