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 ∈
(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
since mx + ny ∈Z.
2. Show that there are arbitrarily long sequences of consecutive integers
primes. In other words, show that given an integer N ≥1, there exists an integer
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.
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
that then r = s by contradiction. If r ≠ s, then switching r, s if necessary, we
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 positivemultiple 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.
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
consequence of theorem.
Induction step . Suppose k ≥ 2 is an integer such that whenever we are given k
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
. 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.
To find gcd(315, 168), we perform the Euclidean algorithm, keeping track of what
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 ·
To double-check, we have 315 = 32 · 5 · 7 and 168 = 23 · 3 · 7, so gcd(315, 168)
= 3 · 7 and
lcm(315, 168) = 23 · 3 · 5 · 7.
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