SOLUTIONS TO HOMEWORK 7
1(a). The text offers a (pictorial) proof that division with remainder works in the
Gaussian integers Z[i]: that if α and β are Gaussian integers with β 6= 0, then there
are Gaussian integers µ and ρ with
α = µβ + ρ,
Norm(ρ) ≤ Norm(β)/2.
(The Gaussian integers µ and ρ may not be unique.) Explain a way (given α and β)
actually to compute µ = m1 + im2 and ρ = r1 + ir2 , using just ordinary arithmetic
operations on the integer coordinates
α = a1 + ia2 ,
β = b1 + ib2 .
If this were a computer-friendly class, I would ask you to write code to compute
m1 , m2 , r1 , and r2 from a1 , a2 , b1 , and b2 . But here some words describing what
to do will suffice. (Hint: By looking at the diagram on page 107, you can perhaps
at least guess that m1 is the integer minimizing
|α − m1 β|2 .
You can find the real number m1 minimizing this expression by geometry.)
Ans: As suggested in the problem, we find m1 by minimizing the expression
|α − n1 β|2 for a real number n1 , and then choosing the integer closest to n1 for
m1 . (We arbitrarily pick one of the two possibilities if n1 is halfway between
two integers.) That distance is minimized when the line connecting n1 β to α is
perpendicular to the line connecting β to the origin. This can be rewritten as
β · (α − n1 β) = 0
(that’s the dot product, where we’re thinking of C as R2 ). This yields the equation
a1 b1 − n1 b21 + a2 b2 − n1 b22 = 0, and so we find m1 is the nearest integer to
a1 b1 + a2 b2
.
b21 + b22
Similarly, for m2 , we find the nearest integer to the real number n2 minimizing
|α − in2 β|2 . We get that m2 is the nearest integer to
a2 b1 − a1 b2
.
b21 + b22
Because ρ = α − µβ, and we now know the coordinates of the three terms on the
right, we can find r1 and r2 .
1(b). As a test of the procedure you developed in (a), find µ and ρ when α = 137,
β = 37 + i. (If you didn’t solve (a), you can still solve this by trial and error.)
Ans: Here a1 = 137, a2 = 0, b1 = 37, b2 = 1, and so m1 is the closest integer to
137 ∗ 37/(372 + 12 ) = 3.7. Hence m1 = 4. Similarly, m2 is the nearest integer to
−.1, i.e., m2 = 0. And ρ = 137 − 4 ∗ (37 + i) = −11 − 4i. As a check, we can
compute the norms:
1
2
SOLUTIONS TO HOMEWORK 7
Norm(β) = 372 + 12 = 1370
Norm(ρ) = (−11)2 + (−4)2 = 137.
1(c). Find gcd(137, 37 + i) in the Gaussian integers. (Hint: it is not 1.)
Ans: We already have the first line of the Euclidean algorithm, namely
137 = 4 ∗ (37 + i) + (−11 − 4i).
We compute the next step by performing the above computation, with α = 37 + i
and β = −11 − 4i. We find µ = −3 + i and ρ = 0. Hence, the greatest common
divisor is −11 − 4i. (This is the same as saying the gcd is 11 + 4i.)
2. In this problem p is a prime congruent to 1 modulo 4.
2(a). Suppose a is any non-zero integer modulo p. Define
b ≡ a(p−1)/4
2
(mod p).
2
Prove that either b ≡ −1 (mod p) or b ≡ 1 (mod p). Prove that if a is a primitive
root modulo p, then the first possibility occurs.
Ans: (b2 )2 ≡ a2∗(p−1)/2 ≡ 1 (mod p) by Fermat’s Little Theorem. Hence, by
the proposition beginning with the word Solutions on page 62 (applied for the
exponent n = 2), b2 ≡ −1 or 1 (mod p). If a is a primitive root, then ax ≡ 1
(mod p) implies p − 1 (the order of a) divides x. Hence, the first possibility occurs
in this case.
2(b). Prove that there is exactly one integer m such that
2 ≤ m ≤ (p − 1)/2,
m2 ≡ −1
(mod p).
Ans: The equation m2 ≡ −1 (mod p) implies that m4 ≡ 1 (mod p) and that m
is not a solution of mn ≡ 1 (mod p) for n < 4. So, we apply the proposition on
page 62 referenced above to conclude that there are at most two such m’s. There
is at least one solution, because p is congruent to 1 (mod 4), and so a(p−1)/4 is one
such m, for a a primitive root. Clearly −m also works. Hence, there is exactly one
in the given range.
2(c). Describe a reasonable way to find the integer m as in (b). (Testing every m
between 2 and (p − 1)/2 is too slow.)
Ans: Let c = 2. The number c(p−1)/4 , call it c0 , is a fourth root of one. It is a
square root of −1 modulo p if c is a primitive root by part (a). So, if (c0 )2 ≡ −1
(mod p), then either c0 or −c0 is the answer. Otherwise, replace c with c + 1, and
repeat the test. This is a reasonable algorithm because there are at least ϕ(p − 1)
primitive roots modulo p as shown in the proof of “Existence of Primitive Roots”
on page 62.
2(d). Suppose that gcd(p, m + i) = x + iy (calculated in the Gaussian integers
Z[i]). Prove that x2 + y 2 = p.
Ans: Because x + iy divides p, Norm(x + iy) divides Norm(p) = p2 . So, Norm(x +
iy) = 1, p, or p2 . If the norm were p2 , and say p = (x + iy)(a + bi), we have
Norm(a + bi) = 1, and hence x + iy is a unit multiple of p. But then x + iy divides
m + i implies p divides m + i, which is a contradiction (since p multiplied by a
SOLUTIONS TO HOMEWORK 7
3
Gaussian integer yields a Gaussian integer whose real and imaginary coefficients
are divisible by p.)
So, we’ve eliminated one of the norm possibilities. It remains to show that p and
m + i are not relatively prime. By the definition of m, we have
(m + i)(m − i) = m2 + 1 ≡ 0
(mod p).
Hence p divides the product (m + i)(m − i) (the first divides the second as integers,
and hence also as Gaussian integers). If p and m + i were relatively prime, then
p would have to divide m − i, which is a contradiction. Hence, Norm(x + iy) =
x2 + y 2 = p.
2(e). Find the integer m in case p = 137.
Ans: We perform the algorithm described in (c).
2136/4 ≡ −1
(mod 137),
so 2 is not a primitive root. But
3136/4 ≡ 100
(mod 137),
2
and so 100 ≡ −1 (mod 137). This isn’t in the range we want though. So m ≡
−100 ≡ 37 (mod 137).
2(f ). Solve the Diophantine equation x2 +y 2 = 137. (This is easy to do by guessing;
but the problems up to now tell you how to write a solution immediately.)
Ans: By 2(d) and 2(e), we are reduced to finding the greatest common divisor of
137 and 37 + i. We already did this in 1(c), and so x = 11 and y = 4 is a solution.
3. This problem is about quadratic algebraic integers (page 125 in the text). Always
N is a fixed non-zero integer not divisible by p2 for any prime p.
3(a).√Suppose that N ≡ 1 (mod 4). If a and b are rational numbers, prove that
a + b N is a quadratic integer if and only if either
a and b are both integers
or
2a and 2b are both odd integers.
√
Ans: Let a + b N satisfy the monic quadratic equation with integer coefficients
x2 + c1 x + c2 . Then
√
√
a2 + b2 N + 2ab N + c1 (a + b N ) + c2 = 0.
√
We can group the coefficients of N together to find that c1 = −2a. Since c1 is an
integer, we have that 2a is an integer, so a = a1 /2 for some integer a1 . We further
have c2 = a2 − N b2 = a21 /4 − N b2 is an integer. Because N is square-free (i.e., for
no prime p does p2 divide N ), N cannot cancel out the square of the denominator
of b, and so b = b1 /2 for some integer b1 . We must have a21 − N b21 is divisible by
4, since c2 is an integer. Thus a21 ≡ N b21 (mod 4). As N ≡ 1 (mod 4), we have
a21 ≡ b21 (mod 4). Hence, either a1 and b1 are both odd, and so 2a and 2b are both
odd, or else a1 and b1 are both even, in which case a and b are both integers.
3(b).√Suppose that N 6≡ 1 (mod 4). If a and b are rational numbers, prove that
a + b N is a quadratic integer if and only if a and b are both integers.
Ans: The proof above goes through as before until we get to the equation a21 ≡
N b21 (mod 4). Now since N ≡ 3 (mod 4), we have a21 ≡ −b21 (mod 4). Since −1 is
4
SOLUTIONS TO HOMEWORK 7
not a square modulo 4, we must have both a1 and b1 are even, and thus both a and
b are integers.
4. This problem is about the quadratic integers
√
√
Z[ −5] = {x + y −5|x, y ∈ Z}.
The norm of such an integer is defined to be
√
Norm(x + y −5) = x2 + 5y 2 ;
this is multiplicative just as for Gaussian integers. The only elements of norm 1
(the units) are ±1. Just as for Gaussian integers, we can define primes and prove
the existence of prime factorization; you can assume all of that.
√
4(a). Prove that there are no elements of Z[ −5] having norm 2 or norm 3.
Ans: An element of norm 2 (resp., 3) would yield integer solutions of the equation
x2 + 5y 2 = 2 (resp., = 3), which do not exist.
√
√
√
4(b). Prove that 2, 3, 1 + −5, and 1 − −5 are all primes in Z[ −5].
Ans: Their norms are 4, 9, 6, 6, respectively. A factorization would have to involve
two terms with norm not equal to one, and thus, since the resulting norms would
multiply to equal 4 or 9 or 6, we would have an element of norm 2 or norm 3, which
we just saw was impossible.
4(c). Using
√ these four primes, give an example showing that unique factorization
fails in Z[ −5].
√
√
Ans: 2 ∗ 3 = (1 + −5)(1 − −5) = 6.