Download Key to Homework 1 August 26, 2005 Math 521 Direction: This

Key to Homework 1
August 26, 2005
Math 521
Direction: This homework is due on September 2, 2005. In order to receive full credit, answer
each problem completely and must show all work.
1. For each value of n listed, find all positive integers less than n and relatively prime to n.
n = 8, 12, 20, 25.
Answer: All integers less than n and relatively prime to n are:
n = 8;
{1, 3, 5, 7}
n = 12,
{1, 5, 7, 11}
n = 20,
{1, 3, 7, 9, 11, 13, 17, 19}
n = 25,
{1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24}
2. Determine the gcd(24 · 32 · 5 · 72 , 2 · 33 · 11) and lcm(24 · 32 · 5 · 72 , 2 · 33 · 11).
Answer: gcd(24 · 32 · 5 · 72 , 2 · 33 · 11) = 2 · 32 = 18.
lcm(24 · 32 · 5 · 72 , 2 · 33 · 11) = 24 · 33 · 5 · 72 · 11 = 1164240.
3. Find integers s and t such that 1 = 7s + 11t. Show that s and t are not unique.
Answer: Let s = 8 and t = −5. Then 7s + 11t = 7(8) + 11(−5) = 56 − 55 = 1.
Similarly let s = −3 and t = 2. Then 7s + 11t = 7(−3) + 11(2) = 21 + 22 = 1.
4. If a and b are integers and n is a positive integer, prove that a mod n = b mod n if and only
if n divides a − b.
Answer: Let a mod n = b mod n = α (say). Then n/(α − a). Similarly n/(b − α). Hence
n/(α − a + b − α). That is n/(b − a).
Suppose n/(b−a). Then using division algorithm, we can write b = nq1 +r1 for some integers q1
and r1 . Similarly a = nq2 + r2 for some integers q2 and r2 . Hence b − a = n(q1 − q2 ) + (r1 − r2 ).
Since n/(b − a), we have r1 − r2 = 0. That is r1 = r2 and hence a mod n = b mod n.
5. Suppose a and b are integers that divide the integer c. If a and b are relatively prime, show
that ab divides c. Show, by example, that if a and b are not relatively prime, then ab need not
divide c.
Answer: a/c implies c = n1 a for some integer n1 . Similarly b/c implies c = n2 b for some
integer n2 . Since gcd(a, b) = 1, there exist integers s and t such that a s + b t = 1. Hence
c = a s c + b t c and therefore c = a s n2 b + b t n1 a = (s n2 + t n1 ) ab. Thus c/ab.
6. Show that 5n + 3 and 7n + 4 are relatively prime for all n.
Answer: Since (5n + 3)(7) + (7n + 4)(−5) = 35n + 21 − 35n − 20 = 1, the integers 5n + 3 and
7n + 4 are relatively prime for any n.
7. Use the Euclidean algorithm to find gcd(34, 126) and write it as a linear combination of 34
and 126.
Answer: gcd (34, 126) = 2 and 2 = gcd(34, 126) = 127(−7) + 34(26).
8. Let p1 , p2 , p3 , ..., pn be primes. Show that p1 p2 · · · pn + 1 is divisible by none of these primes.
Answer: Suppose not. Then there exists a p ∈ {p1 , p2 , p3 , ..., pn } such that p/(p1 p2 · · · pn + 1).
Since p/p1 p2 · · · pn therefore we have p/1 which is a contradiction.
9. Prove that for every integer n, n3 mod 6 = n mod 6.
Answer: To prove n3 mod 6 = n mod 6, we must show that 6/(n3 −n), that is 6/(n−1)n(n+1).
For this we need to consider 6 cases.
Case 1. Suppose n + 1 = 0 mod 6. Then in this case, 6/(n − 1)n(n + 1).
Case 2. Suppose n + 1 = 1 mod 6. Then 6/n and hence 6/(n − 1)n(n + 1).
Case 3. Suppose n + 1 = 2 mod 6. Then 6/(n − 1) and hence 6/(n − 1)n(n + 1).
Case 4. Suppose n + 1 = 3 mod 6. Then 6/(n − 2). This implies that 3/(n − 2). Therefore
3/(n + 1). Now consider two sub-cases. Suppose n + 1 is an odd integer. Then n is an even
integer and this fact and 3/(n + 1) implies that 6/(n − 1)n(n + 1). Next suppose n + 1 is an
even integer. Since n + 1 is even and 3/(n + 1), we have 6/(n + 1) and hence 6/(n − 1)n(n + 1).
Case 5. Suppose n + 1 = 4 mod 6. Then 6/(n − 3). This implies that 3/(n − 3). Therefore
3/n. Now consider two sub-cases. Suppose n + 1 is odd. Then n must be even and hence
hence 6/(n − 1)n(n + 1). Next suppose n + 1 is even. Since n + 1 is even and 3/n we have
6/(n − 1)n(n + 1).
Case 6. Suppose n + 1 = 5 mod 6. Then 6/(n − 4). This implies that 3/(n − 1). Now consider
two sub-cases. Suppose n + 1 is odd. Then n is an even integer. Since n is even and 3/(n − 1),
we get 6/(n − 1)n(n + 1). Suppose n + 1 is even. Then clearly 6/(n − 1)n(n + 1).
10. Solve the equation x2 = 4 mod 5 on the set of integers.
Answer: x2 = 4 mod 5 implies 5/(x2 − 4). That is 5/(x − 2)(x + 2). Then by Euclid’s lemma
5/(x − 2) or 5/(x + 2). That is x = 2 mod 5 or x = −2 mod 5. Hence the solution set is
{x | x = 5k + 2 or x = 5k − 2 for all k ∈ Z} which is
{ ±2, ±3, ±7, ±8, ±12, ±13, ±17, ....}
.