# SOLUTIONS TO HOMEWORK 2 Due Wednesday, September 21 in class. ```SOLUTIONS TO HOMEWORK 2
Due Wednesday, September 21 in class.
1(a) Use the Euclidean algorithm to find gcd(9797, 1649).
Ans: We use division with remainder to find the gcd, as on p. 25.
9797 = 5 ∗ 1649 + 1552
1649 = 1 ∗ 1552 + 97
1552 = 16 ∗ 97 + 0
Hence gcd(9797, 1649) = 97.
1(b). Find integers m and n so that
gcd(9797, 1649) = m &middot; 9797 + n &middot; 1649.
Ans: One method is given in the example on p. 27:
(9797, 1649) = (a, b)
(1649, 1552) = (b, a − 5b)
(1552, 97) = (a − 5b, b − (a − 5b))
From this we can read off 97 = −1 ∗ 9797 + 6 ∗ 1649. So, in the notation above, one
possible solution is (m, n) = (−1, 6).
1(c). Write the continued fraction for 9797/1649 (see page 26 of the text).
Ans: Here we follow pp. 25-26, using our three lines of computation from part
(a):
5 ∗ 1649 + 1552
1552
1
9797
=
=5+
=5+
1649
1649
1649
1649/1552
1
1
1
=5+
=5+
1
1649/1552
(1 ∗ 1552 + 97)/1552
1 + 16
√
2. Let R be the collection of complex numbers m + n −3, with m and n integers.
I’ll write assumptions like this as
√
R = {m + n −3 | m, n ∈ Z}.
5+
2(a). Explain why R is closed under addition and multiplication.
√
√
Ans: Let r1 = m1 + n1 −3 and r2 = m2 + n2 −3 denote two arbitrary elements
of R. We need to check that r1 + r2 ∈ R and r1 r2 ∈ R to show that R is closed
under addition and multiplication, respectively. First,
√
r1 + r2 = (m1 + m2 ) + (n1 + n2 ) −3,
and since m1 + m2 ∈ Z and n1 + n2 ∈ Z, we have that r1 + r2 ∈ R. Hence, R is
1
2
SOLUTIONS TO HOMEWORK 2
Now we compute r1 r2 :
√
√
(m1 + n1 −3)(m2 + n2 −3) =
√
√ √
= (m1 m2 + n1 n2 −3 −3) + (m1 n2 + m2 n1 ) −3
√
= (m1 m2 − 3n1 n2 ) + (m1 n2 + m2 n1 ) −3.
Because m1 m2 − 3n1 n2 ∈ Z and (m1 n2 + m2 n1 ) ∈ Z, we have that r1 r2 ∈ R, and
so R is closed under multiplication.
2(b). Define a “norm” on R by
√
km + n −3k = m2 + 3n2 .
Prove that krk is a non-negative integer (for all r ∈ R), and that
kr &middot; sk = krk &middot; ksk
(r, s ∈ R).
Ans: Because m, n ∈ Z by the definition of R, m2 and n2 are non-negative
integers, and hence m2 + 3n2 is a non-negative integer.
We’ll now use our calculation of r1 r2 above to prove
kr1 &middot; r2 k = kr1 k &middot; kr2 k
(r1 , r2 ∈ R).
On one hand,
kr1 &middot; r2 k = (m1 m2 − 3n1 n2 )2 + 3 ∗ (m1 n2 + m2 n1 )2
= (m21 m22 + 9n21 n22 − 6m1 m2 n1 n2 ) + 3(m21 n22 + m22 n21 + 2m1 m2 n1 n2 )
= m21 m22 + 9n21 n22 + 3m21 n22 + 3m22 n21
On the other hand,
kr1 k &middot; kr2 k = (m21 + 3n21 )(m22 + 3n22 )
= m21 m22 + 9n21 n22 + 3m21 n22 + 3m22 n21 .
This proves the desired result.
Note: If you know some results from complex variables, there is a much cleaner
solution. (The proofs of the two necessary facts are similar to the above proof:
you write things out in terms of real and complex coordinates and expand.) Let z
denote the complex conjugate of z. Here are the facts: for every x, y ∈ C,
(1) xy = x ∗ y
(2) xy = yx.
Now, note that kxk = x ∗ x. Then using the two facts above,
kxyk = xyxy = xxyy = kxkkyk.
2(c). Show that the only elements of R having a multiplicative inverse are &plusmn;1.
Ans: First, &plusmn;1 both have multiplicative
inverses: each is its own inverse. Now,
√
consider any number r = m + n −3 with a multiplicative inverse r−1 . By part
(b), we know k1k = kr ∗ r−1 k = krk ∗ kr−1 k. Since k1k = 1, we have
1 = krk ∗ kr−1 k.
Also by part (b), we know that krk and kr−1 k are non-negative integers. But if the
product of two non-negative integers is 1, then both of those non-negative integers
are 1. The equation krk = 1 is equivalent to m2 + 3n2 = 1. Because m and n are
SOLUTIONS TO HOMEWORK 2
3
integers, it’s clear that m = &plusmn;1, n = 0. So, r = &plusmn;1. Since our only restraint on r
was that it have norm 1, we’re done.
2(d). Call an element r√of R prime if√it has exactly four divisors (namely &plusmn;1 and
&plusmn;r). Prove that 2, 1 + −3, and 1 − −3 are all prime in R.
Ans: The idea is similar to the idea in the previous problem. We write a factorization r = r1 r2 . We then compute norms of both sides, and get that r1 or r2 is
&plusmn;1, and hence the only divisors are the listed ones. Here we go.
Assume 2 = r1 r2 . Then k2k = kr1 k ∗ kr2 k. And we compute k2k = 4. Again,
since norms are non-negative integers, we have kr1 k, kr2 k have possible values 1, 2,
and 4, and their product must be 4. But m2 + 3n2 = 2 has no integer solutions.
Hence one of kr1 k, kr2 k is 1, and the other is 4. From part (c), the only elements
with norm 1 are &plusmn;1. If it’s 1, then the other element is 2. If it’s −1, the other
element is −2.
√
The
√ exact same argument works for the other two cases, because k1 + −3k =
k1 + −3k = k2k = 4. Hence, these three elements are all prime.
2(e). Prove that any element of R other than 0 and &plusmn;1 is a product of primes in
R: so prime factorization is possible in R.
Ans: For any non-zero element r ∈ R, clearly all of 1, −1, r, −r are divisors of it.
If r 6= 0, 1, −1, then the four given elements are distinct. So, an element r 6= 0, 1, −1
which is not prime has more than 4 divisors.
We want to show that every r 6= 0, 1, −1 has a prime factorization. We’ll prove
this by induction on the norm of r. As we saw above, given r 6= 0, 1, −1, norms
0, 1, 2 √
are impossible. If krk = 3, then from the formula for the norm it’s clear that
r = &plusmn; 3. A factorization r = r1 r2 leads to a factorization kr1 k ∗ kr2 k = krk = 3,
and as in the previous problem, since norms are non-negative integers, one of r1
and r2 has norm 1 (this is true in general for any element r with prime norm [prime
as an integer, that is]). Hence one of them is &plusmn;1, and hence the other is &plusmn;r.
So, the base case for our induction is complete: if krk = 3, then r is a product
of primes; in fact, r is prime in this case. For our inductive hypothesis, we now
assume that for any r 6= 0, 1, −1 with krk &lt; N (where N ≥ 4), r is a product of
primes. Now consider an r with norm N . If r = r1 r2 only has solutions with either
r1 or r2 equalling &plusmn;1, then r is prime and we’re done. Otherwise both kr1 k and
kr2 k ≥ 2. Since kr1 k ∗ kr2 k = N , we have 2 ≤ kr1 k, kr2 k &lt; N . We can then apply
the inductive hypothesis to write both r1 and r2 as a product of primes, and thus
also r = r1 r2 as a product of primes.
2(f ). What remark would you make about the equations
√
√
2 &middot; 2 = 4 = (1 + −3)(1 − −3)?
Ans: They show√that the factorization
into primes is not unique, since we saw
√
above that 2, 1 + −3 and 1 − −3 were all primes.
3. Suppose that a &gt; b &gt; 1 are relatively prime natural numbers. According to the
Euclidean algorithm, it is possible to find integers x and y so that
ax + by = 1.
Prove that we can actually arrange
0 &lt; x &lt; b,
−a &lt; y &lt; 0.
4
SOLUTIONS TO HOMEWORK 2
(Hint: page 34 of the text might help.)
Note the correction to the originally posted homework: we now have
a &gt; b &gt; 1 rather than a &gt; b ≥ 1.
Ans: Because a(−bq) + b(aq) = 0 for any integer q, the given equation ax + by = 1
implies a(x − bq) + b(y + aq) = 1. Applying division with remainder to x and b,
we find q and r so that x = bq + r with 0 ≤ r &lt; b. Then ar + b(y + aq) = 1. This
r satisfies the required inequality, because if r = 0, then we see that b divides 1,
which is impossible since we require b &gt; 1.
Set s = y + aq. We claim that −a &lt; s &lt; 0. Assume first s ≤ −a. Then
1 = ar + bs ≤ ar − ba = a(r − b) &lt; 0,
where the last inequality holds because r &lt; b and a is positive. But we see 1 &lt; 0,
which is a contradiction. Now assume s ≥ 0. Then
1 = ar + bs ≥ ar ≥ a &gt; 1,
4. Problems 2.5.6 and 2.5.7 on page 33 of the text.
Ans: 2.5.6: Note that q is odd, and thus is a prime different from 2. By the
uniqueness of prime factorization, we know any divisor of 2p−1 q (which we see there
written in terms of its prime factorization) has the form 2a q b where 0 ≤ a ≤ p − 1
and 0 ≤ b ≤ 1 are integers. We are only interested in the proper divisors, and so
we remove 2p−1 q 1 from consideration. We are left precisely with the given list.
Ans: 2.5.7 The equality is more evident when we write the left-hand side as
(2 − 1)(1 + 2 + 22 + &middot; &middot; &middot; + 2p−2 ).
Expanding, we see that all the terms cancel except for −1 and 2p−1 . Hence
1 + 2 + 22 + &middot; &middot; &middot; + 2p−2 = 2p−1 − 1.
Now we add up all the proper divisors:
1 + 2 + &middot; &middot; &middot; + 2p−1 + q(1 + 2 + &middot; &middot; &middot; + 2p−2 ) =
(1 + 2 + &middot; &middot; &middot; + 2p−2 ) + 2p−1 + q(1 + 2 + &middot; &middot; &middot; + 2p−2 ) =
(2p−1 − 1) + 2p−1 + q(2p−1 − 1) =
2p − 1 + (2p − 1)(2p−1 − 1) =
2p − 1 − (2p − 1) + q2p−1 = 2p−1 q.
```