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$p,q \in \mathbb{P}$, primes

For $x \not\equiv 0 \bmod p$ you can write $(x-1)(x-2) \dots (x - (p-1)) \equiv 0$ mod $p$

Is there a way to do the same for a a composite modulus $pq$?

Note: $(x-1)(x-2) \dots (x-p) \dots (x-q) \dots (x - (pq-1)) \equiv 0$ mod $pq$ does not work. If $x \equiv 0$ mod $pq$ then the terms $(x-p)$ and $(x-q)$ become $pq$ which makes the polynomial $\equiv 0$ mod $pq$.

Also $x^{(p-1)(q-1)} \equiv 1$ mod $pq$ if $\gcd(x,pq) = 1$ does not cover the case when $x = kp$ or $x = kq$.

------- to clarify ------------------------------------------

Find a polynomial modulo pq such that $x \equiv 0 $ (mod pq) is not a solution but all other numbers are solutions. i.e. a filter.

This can be done if the modulus is a single prime.

@André Nicolas demonstrated that it cannot be done for a composite modulus.

  • Your one-variable polynomial is $x^{p-1}-1\equiv0\pmod p$ – Lubin Jul 18 '16 at 13:59
  • _@Lubin - if $p = 3$ and $q = 5$ Let $x = 3 \not\equiv 0 $ (mod $15$). $3^2 \equiv 0 \not\equiv 1$ (mod $3$) –  Jul 19 '16 at 01:41
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    Sorry, I did not claim to be making a contribution to the two-prime question, rather I was just saying that instead of your large product in line 2, you could have used the form I pointed out. – Lubin Jul 19 '16 at 12:35

1 Answers1

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Let $f(x)$ be a polynomial. Suppose that the distinct primes $p$ and $q$ are roots of $f(x)\equiv 0\pmod{pq}$.

Since $(x-p)(x-q)$ is monic, there is a polynomial $g(x)$ such that $f(x)=(x-p)(x-q)g(x)+ax+b$ for some integers $a$ and $b$.

Since $p$ and $q$ are roots of $f(x)\equiv 0\pmod{pq}$, it follows that $pq$ divides $ap+b$ and $pq$ divides $aq+b$. So $p$ and $q$ both divide $b$, and therefore $p$ and $q$ both divide $a$.

It follows that $pq$ divides $f(pq)$.

Remark: In hindsight, there is a one line proof. Since $f(0)\equiv 0\pmod{p}$ and $f(0)\equiv 0\pmod{pq}$, it follows that $f(0)\equiv 0\pmod{pq}$. But I might as well keep my ugly argument above: it is what I first wrote down.

André Nicolas
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