1. Suppose a, b, c ∈ Z.

(a) Show that if a|b and c ≠ 0, then ca|cb.

If a|b, then ax = b for some x ∈ Z, so cax = cb so (ca)x = (cb) i.e. ca|cb.

(b) Show that if a|b and b|c, then a|c.

If ax = b and by = c with x, y ∈ Z, then (ax)y = c = a(xy) so a|c since xy ∈
Z.

(c) Show that if a|b and a|c, then a|(mb + nc) for all m, n ∈ Z.

If ax = b and ay = c with x, y ∈ Z, then mb+nc = max+nay = a(mx+ny) so
a|(mb+nc)

since mx + ny ∈Z.

2. Show that there are arbitrarily long sequences of consecutive integers
containing no

primes. In other words, show that given an integer N ≥1, there exists an integer
a such

that a + 1, a + 2, . . . , a + N are all composites. Hint: try a = (N + 1)! + 1.
Look for an

“obvious” divisor of a + 1, an “obvious” divisor of a + 2 etc.

With a = (N + 1)! + 1, we have 2|2 and 2|(N + 1)! so 2|(a + 1) by Problem 1.
Indeed, for

2 ≤j ≤N + 1, (N + 1)! + j = a + (j − 1) is divisible by j because
j|j and j|(N + 1)!. On

the other hand, clearly each such j satisfies 2 ≤ j < a + (j − 1) so a + (j
− 1) cannot be a

prime, hence must be composite.

3. Suppose a, b, n are integers, n ≥1 and a = nd + r, b = ne + s with 0 ≤ r,
s < n, so

that r, s are the remainders for a ÷ n and b ÷ n, respectively. Show that r = s
if and only if

n|(a − b). [In other words, two integers give the same remainder when divided by
n if and

only if their difference is divisible by n.]

Suppose r = s. Then r = s = a − nd = b − ne. Rearranging the last two
equalities , we

get a − b = nd − ne = n(d − e) so n|(a − b). Conversely, suppose n|(a − b); we
will prove

that then r = s by contradiction. If r ≠ s, then switching r, s if necessary, we
can assume

without loss of generality that r > s. By assumption, n|(a − b). Thus, nx = a −
b for some

x ∈ Z so

a − b = nd + r − ne − s = n(d − e) + r − s = nx.

Rearranging the last equality we have r − s = n(d − e − x)
and d − e − x ∈ Z so n|(r − s).

Since r > s, we conclude that r − s ≥ n because the least positive multiple of n
is n itself .

But we have 0 ≤ s < r < n so r − s < n, a contradiction. We have thus shown that r
= s if

n|(a − b).

4. If n ≥2 and are n integers whose
product is divisibe by p, then at

least one of these integers is divisible by p, i.e.
implies that
then there exists

1≤ j ≤ n such that . Hint: use induction on n.

Proof by induction on n. Base case n = 2 was proved in class and in the notes as
a

consequence of theorem.

Induction step . Suppose k ≥ 2 is an integer such that whenever we are given k
integers

whose product is divisible by p (i.e.
)), there exists

1≤ j ≤k such that p|mj . Now suppose we are given k + 1 integers
such that

. We have
where . By the base case, we conclude

that either p|m or . If
, then certainly there exists 1≤ j≤
k + 1 such that

, namely j = k + 1. Otherwise, p|m and by
the induction hypothesis, then there exists

1≤ j ≤ k such that . Thus, we have shown that there exists 1 ≤j ≤k
+ 1 such that

, completing the induction step . By PMI, we are done.

5. (a) Calculate gcd (315, 168) using the Euclidean algorithm, then use this
information

to calculate lcm (315, 168). Determine integers x, y such that 315x + 168y =
gcd(315, 168).

You may use the Blankinship version of the Bezout algorithm if you wish. Now
obtain the

prime factorizations of 315 and 168 to double -check your computation of the gcd
and lcm of

315 and 168.

To find gcd(315, 168), we perform the Euclidean algorithm, keeping track of what
it does

to the two extra columns comprising an “identity” matrix.

We read off that gcd(315, 168) = 21 (the last non- zero
remainder ) and that −1(315) +

2(168) = 21. We also have 8(315)−15(168) = 0 i.e. 8(315) = 15(168) and that
lcm (315, 168) =

8(315) = 15(168) = 2520. Or we could use lcm(315, 168) gcd(315, 168) = 315 ·
168.

To double-check, we have 315 = 3^{2} · 5 · 7 and 168 = 2^{3} · 3 · 7, so gcd(315, 168)
= 3 · 7 and

lcm(315, 168) = 2^{3} · 3 · 5 · 7.

(b) Calculate gcd (89, 148) using the Euclidean algorithm.

Thus, gcd(148, 89) = 1.