Download SOLUTIONS TO MATH 38B GRADED HOMEWORK 1 Exercise 1

SOLUTIONS TO MATH 38B GRADED HOMEWORK 1
Exercise 1
Show that for every natural number k, the number 2k + 1 and 9k + 4 are relatively prime.
Solution. If k = 0 then we have gcd(1, 4) = 1. For k = 1 we get gcd(13, 3) = 1. For k ≥ 2 it follows
from the Euclidean algorithm. We have
9k + 4 = 4(2k + 1) + k
2k + 1 = 2(k) + 1.
The remainder k in line the first step is valid because k < 2k + 1 and the remainder 1 in the second
step is valid because 1 < k. It follows that gcd(2k + 1, 9k + 4) = 1.
Exercise 2
Determine, as a function of n, the number gcd(n + 1, n2 + 1) for every natural number n.
Solution. We claim that
(
1 if n is even,
2
gcd(n + 1, n + 1) =
2 if n is odd.
If n = 1 then we have gcd(2, 2) = 2 so our claim is true in this case. For n ≥ 2 we see this by the
Euclidean algorithm. Observe that (n + 1)(n − 1) = n2 − 1 so that
n2 + 1 = (n − 1)(n + 1) + 2.
This is a valid first step in the Euclidean algorithm beacuse 2 < n + 1. Thus we see that for n ≥ 2,
gcd(n + 1, n2 + 1) = gcd(2, n + 1). For n odd, n + 1 is even and hence we get 2. For n even, n + 1
is odd and hence we get 1.
Exercise 3
Let a and b be different natural numbers. Show that there exist infinitely many natural numbers n
such that a + n and b + n are relatively prime.
Solution. We deduce the statement using the infinitude of primes. Without loss of generality
suppose that a < b. If n is an integer such that b + n is prime then since 1 < a + n < b + n, it follows
that gcd(a + n, b + n) = 1. Since there are infinitely many primes p > b there are infinitely many n
such that b + n is prime and we are done by the second sentence.
No one solved it this way, but I must admit the first solution that came to mind for me was to
explicitly construct the n as follows. Suppose that a = 0 and b > 0; the general case is not much
harder. Then, for each n > b the division algorithm says that gcd(n, b + n) = gcd(n, b). From this,
it now suffices to show that there are infinitely many n such that n is coprime to b. But, this is
obvious by taking integers of the form kb + 1.
Exercise 4
Let n > 1 be a natural number with the following property: for each pair of natural numbers a and
b dividing n such that a > 1 and b > 1, one has gcd(a, b) > 1. What can you say about n?
Solution. We claim n is a non-zero power of a prime number p. If n = pr where r ≥ 1 then a and b
are chosen from among p, p2 , . . . , pr . Clearly any pair of choices leads to gcd(a, b) > 1. Conversely,
suppose that there exists two different prime divisors p and q of n. Take a = p and b = q and since
p 6= q and both are prime we have gcd(a, b) = 1.
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