1. If n E Z n>1 . N is prime if and only if (n-2)!= 1 modn
2. Let n be composite and >4. Prove (n-1)! = 0mod n
3. Show that 11 divides 456^654 +123^321
4. Find the solution to 9x=21mod23
Thanks for any help!
If n is prime then (n-1)!=-1 (n). Write as (n-1)(n-2)!=-1 (n). But n-1=-1 (n). Thus, we are left with after dividing (n-2)!=1 (n).
See this.2. Let n be composite and >4. Prove (n-1)! = 0mod n
Find 456^654 (mod 11) and 123^321 (mod 11). Now so their remainders add up to a multiple of 11.3. Show that 11 divides 456^654 +123^321
Note gcd(9,23)=1 and 1|21 so there is a unique solution.4. Find the solution to 9x=21mod23
Divide by 3 to get, 3x=7(mod 23) which is equvailent to 3x=30(mod 23).
Thus, we get x=10(mod 23).