Answer:
Check the explanation
Step-by-step explanation:
(a)Let p be the smallest prime divisor of (n!)^2+1 if p<=n then p|n! Hence p can not divide (n!)^2+1. Hence p>n
(b) (n!)^2=-1 mod p now by format theorem (n!)^(p-1)= 1 mod p ( as p doesn't divide (n!)^2)
Hence (-1)^(p-1)/2= 1 mod p hence [ as p-1/2 is an integer] and hence( p-1)/2 is even number hence p is of the form 4k+1
(C) now let p be the largest prime of the form 4k+1 consider x= (p!)^2+1 . Let q be the smallest prime dividing x . By the previous exercises q> p and q is also of the form 4k+1 hence contradiction. Hence P_1 is infinite
1) 12
2) 2y
3) 12+2y
4) 12+2y
Answer: First answer: hexagon
second answer: triangle
Step-by-step explanation: right on edge 2020
Common ratio = 8/2 = 32/8 = 4
10th term = a1*r^(n - 1) where a1 = 2 , r = 4 and n = 10
= 2 * 4^9
= 524,288
Answer:
y = -14 in simplest form hope this helped
Step-by-step explanation:
