Can x mod (N - a) or x mod (N + a) be calculated just by knowing x mod N? where a is an arbitrary integer, and N is a prime. e.g. i know 22! mod 23 ā” 22 using Wilson's theorem. lets say i want to know 22! mod 24, is there a way i can do so without solving the factorial?
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I added an answer which discusses the general case. ā Bill Dubuque Jan 08 '19 at 20:21
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If $\,d := \gcd(n,m)=1\,$ then there is no relationship between $\,a = x\bmod n\,$ and $\, b = x\bmod m\,$ since CRT $\,\Rightarrow\,x\equiv a\pmod{\!n},\ x\equiv b\pmod{\! m}\ $ is solvable for all values of $\,a,b.\,$ This is true in your example $\iff 1 = \gcd(n,n\pm a) = \gcd(n,a).$
However, when $\,d > 1\,$a solution exists iff $\,b\equiv a\pmod{\!d},\,$ and this does place constraints on the values $\,b\,$ that $\,x\,$ can take $\!\bmod m$. Then $\,b\,$ is determined uniquely iff $\,m\mid d\iff m\mid n\ $ (i.e. iff $\,n\pm a\mid n\,$ in your example), hence $\,x\bmod m = (x\bmod n)\bmod m\,$ is uniquely determined. This arises frequently, e.g. knowing $\, u = x\bmod 10 = $ units digit of $x$ we can compute its parity by taking the parity of its units digit, i.e. $\ x\bmod 2 = (x\bmod 10)\bmod 2 = u\bmod 2,\,$ e.g. $\,123\bmod 2 = 3\bmod 2 = 1$.
Bill Dubuque
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For the specific case $22!$, the answer is easy: since $4 \leq 22$ and $6 \leq 22$ you have that $22! = 0 \pmod{24}$. However, we didn't use the fact we got from Wilson's Theorem, and it doesn't help us in general.
The answer to your question is no, and here is an illustrative example: suppose you know that $x = 5 \pmod{10}$. Then you could have, among many options, that $x = 5$, or that $x = 15$. In the former case, $x = 5 \pmod{12}$. In the latter case, $x = 3 \pmod{12}$.
Mees de Vries
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1But the answer can be yes too, e.g. $,n=10, nāa=5.\ $ See my answer for the general case. ā Bill Dubuque Jan 08 '19 at 20:24