0

Prove that 97^104 - 1 divisible by 105

So you do 105 = 3 * 5 * 7

then prove that 97^104 = 1 mod 3, 97^104 = 1 mod 5, 97^104 = 1 mod 7

and then you can combine the moduli since there aren't any common factors to get 97^104 = 1 mod 3 * 5 * 7 = 1 mod 105

hence proving that it is divisible

my question is why is this:

1.) why is it okay to split a number into its factors and then check for divisibility by each one of them and why does this prove that the number is divisible by it product of the factors

2.) why is it okay to combine the moduli in the last step and what happens if you want to combine the moduli and they share common factors?

Bill Dubuque
  • 272,048
JerSci
  • 49
  • more generally, show that $a$ divides $c$ and $b$ divides $c$, $\gcd (a,b)=1$ implies $a b$ divides $c$ – Lozenges May 04 '21 at 08:51
  • By CCRT in dupe: $,x\equiv 1\bmod m$ & $!\bmod n$ & $!\bmod k$ $\iff x\equiv 1\pmod{!\ell},\ \ell = {\rm lcm}(m,n,k)\ \ $ – Bill Dubuque May 04 '21 at 08:51

0 Answers0