where 
is an integer. Hence since d divides b, d also divides qb = m. Thus d divides m.
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3. This is false, try a = 3, b = 2 and c = 7 as one example.
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1-2-8. Try the Euclidean Algorithm.
9. For any two integers a > 0, b > 0, let m = lcm(a, b) and d = gcd(a, b). Then we know that md = ab and so, where is an integer. Hence since d divides b, d also divides qb = m. Thus d divides m.
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3. (a). Suppose that p is a prime and p = 3m + 1 for some integer m. Then first note that p is not equal to 2 (since no integer m can satisfy 3m + 1 = 2). But now, m must be even (as otherwise p would be even), and so m = 2k for some integer k and hence, p = 6k +1.
(b). First recall (or verify as an exercise) that the product of any collection of integers of the form 3k+1 is also of the form 3k+1). Now suppose that n is of the form 3n + 2. Then n is not divisible by 3 and so all its factors must be of the form 3k + 1 or 3k + 2. But if every prime factor of n were of the form 3k+1, then since the product of integers of the form 3k+1 is also of the form 3k+1, then n would be of the from 3k+1. But that's a contradiction.
(c). Suppose p is of the form 
. Then since 
, both of these last two factors are larger than 1 for n >= 3, and so the only value of n that could be a prime is n = 2 and that gives p = 7.
(d). Suppose that p is a prime and 3p + 1 is a perfect square. Then 
. Hence by the Fundamental Theorem of Arithmetic, n+1 and n-1 must be 3 and p in some order. So either p = n-1 and 3 = n+1 which forces n=2 and p=1 or p=n+1 and 3=n-1 which forces, n = 4 and p = 5. So the only possible value for p is 5.
(e). Suppose that p is a prime and p is of the form 
. Then since both of the factors n+2 and n-2 are integers larger than 1 whenever n>=4, we must have 0<=n<=3 and the only possible value of p is 5 (when n=3).
4. Suppose that p >= 5 is a prime. Then p is either of the form 6k+1 or 6k+5 (since all the other possible forms are 6k, 6k+2, 6k+3, or 6k+4 and each of these is divisible by 2 or 3).
Suppose that p is of the form 6k+1. Then 
which is divisible by 3.
On the other hand, if p is 6k+5, for some integer k, then 
which is also divisible by 3.
5. (a). First note that by the Fundamental Theorem of Arithmetic, 
and 
have exactly the same prime divisors. If p is a prime divisor of 
, then p must divide 
. Thus 
for some integer m, and so 
and hence 
.
6. (a). Done in class - recall that 
.
(b). If n > 4 and composite then n = ab for some two integers n>a >= b > 2. But then if ab, we would have (n-1)! = (n-1)(n-2)...a...b...1, and so certainly n is divisible by both a and b. If a=b, then (n-1)! = (n-1)(n-2)...2a...a...1.
(c). Since, 
, letting 
, we get 
.
12. 
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5. If n is a 3-digit number, then n less than 1000. So we need only try prime divisors
. So the largest prime that must be tried is 31.
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9(a). Of n, n+2, and n + 4 one of these must be a multiple of 3. Just consider the three cases n is of the form 3k, 3k+1 or 3k+2.
10. For each 2 <= k <= n+1, (n+1)! - k is divisible by k and hence is composite.
22. Hint: Show that if p is a prime that divides both 
then p divides 13.
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4 (a). 
4 (b).
, so the remainder is 6 (since 
)
16. Hint: Check that 
. So the remainder is 1.