Solutions
Math 4400
HOMEWORK #1
1. Let m be the least common multiple of two positive integers a, b. If M is any other common
multiple of a and b, prove that m divides M . (Hint: use division with remainder of M by m.)
Since m is the least common multiple, then m ≤ M . The division algorithm yields
M = qm + r
where
0 ≤ r < m.
Since m and M are both multiples of a, then so is r = M − qm. Likewise, r = M − qm
is a multiple of b. Hence, r is a common multiple of a and b. But m is the least common
multiple, so the only possibility is r = 0. Therefore, M = qm, as desired.
2. Let d be the greatest common divisor of two positive integers a, b. If D is any other common
divisor of a and b, prove that D divides d. (Hint: use the Fundamental Theorem of Arithmetic.)
Since d = gcd(a, b), by the Fundamental Theorem of Arithmetic, there exist integers
x, y such that
ax + by = d.
Since D divides both a and b, D must divide ax + by, i.e., D must divide d.
3. Let a, b, x be positive integers such that gcd(a, x) = 1 and x divides the product ab. Prove that
x divides b. (Hint: use the Fundamental Theorem of Arithmetic.)
Since gcd(a, x) = 1, there exist integers u, v such that
au + xv = 1.
Multiply both sides by b to get
abu + xbv = b.
By assumption, x divides ab, so x also divides abu + xbv, i.e., x divides b.
4. For any two positive integers a, b, prove that lcm(a, b) · gcd(a, b) = ab. (Hint: consider first the
case gcd(a, b) = 1.)
For simplicity, let d = gcd(a, b) and m = lcm(a, b). We are trying to prove that
md = ab.
ab
First, we show that md divides ab. Let n = . This is an integer that can be expressed
d
in the following two ways.
a
b
n = · b and n = a · .
d
d
Therefore, n is a multiple of a and n is a multiple of b. From the first problem, since
n is a common multiple of both a and b, then m must divide n. Thus,
ab
md divides nd =
· d = ab.
d
1
Now, we show that ab divides md. By the Fundamental Theorem of Arithmetic, there
exist integers x, y such that
ax + by = d.
Multiply both sides by m to get
max + mby = md.
Now, m is a multiple of b, so max is a multiple of ab. Similarly, m is a multiple of a,
so mby is a multiple of ab. Hence, max + mby is a multiple of ab, i.e., md is a multiple
of ab.
Therefore, since md divides ab and ab dividies md and these are all positive integers,
it must be true that md = ab.
5. Let f (x) be a polynomial with integer coefficients, say
f (x) = cn xn + cn−1 xn−1 + · · · + c2 x2 + c1 x + c0 ,
and suppose that r = a/b is a rational zero of f (x),i.e., f (r) = 0. Prove that b divides cn and
that a divides c0 . (This is the crux of the rational zeroes test.)
It may be assumed that a/b is in reduced form. Plugging in r = a/b to f (x), we get
a n−1
a 2
a
a n
+ cn−1
+ · · · + c2
+ c1
+ c0 .
0 = cn
b
b
b
b
To clear denominators, we multiply both sides by bn .
0 = cn an + cn−1 an−1 b + · · · + c2 a2 bn−2 + c1 abn−1 + c0 bn .
We can write
cn an = −b cn−1 an−1 + · · · + c2 a2 bn−1 + c1 abn−2 + c0 bn−1 ,
so b divides cn an . Since gcd(a, b) = 1, so b must divide cn (see problem #3).
Similarly, we can write
c0 bn = −a cn an−1 + cn−1 an−2 b + · · · + c + 2abn−2 + c1 bn−1 ,
so a divides c0 bn . Since gcd(a, b) = 1, it must be true that a divides c0 .
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