M3P14 EXAMPLE SHEET 3 SOLUTIONS
1. Give the prime factorizations, in Z[i], of the following elements of Z[i].
Be sure to justify that each of the factors is prime!
1a. 85
We have 85 = 17 ∗ 5 = (4 + i)(4 − i)(2 + i)(2 − i); each of these factors is
prime because its norm is prime.
1b. 18 + 13i
Note that N (18 + 13i) = 493 = 17 ∗ 29, so any prime divisor of 18 + 13i
divides 17 or 29. The possible primes are thus associates of 4 + i, 4 − i,
5 + 2i, and 5 − 2i, and one checks that (4 + i)(5 + 2i) = 18 + 13i.
1c. 7 − i
The norm of 7 − i is 50, so any prime divisor divides 2 or 5. The possible
prime divisors are thus associates of 1 + i, 2 + i, or 2 − i. In fact we have
7 − i = (2 − i)2 (1 + i).
2. Find a greatest common divisor, in Z[i], of the following elements of Z[i]:
2a. 65 and 14 + 21i
One can do this via the Euclidean algorithm, but it is easier here to simply
factor both numbers. We have 65 = 13 ∗ 5 = (3 + 2i)(3 − 2i)(2 + i)(2 − i).
We also have 14 + 21i = 7(2 + 3i). Since 2 + 3i is an associate of 3 + 2i, it is
a common divisor of 65 and 14 + 21i; the factorizations show it is a greatest
common divisor.
2b. 52 and 8 + i
We have 8 + i = (3 + 2i)(2 + i), so we must check which of these divides
52. It is clear that (3 + 2i) does, since 3 + 2i divides 13 and hence 52. On
the other hand 2 + i does not, as its norm is 5 which is relatively prime to
52. Thus the GCD is 3 + 2i.
3a. Let n be an integer. Show that if 4n is the sum of three squares, then so
is n. [HINT: if 4n = a2 + b2 + c2 , show that all of a, b, and c must be even.]
Reduce mod 4; we have a2 + b2 + c2 ≡ 0 (mod 4). Since any square mod 4
is zero or 1, the only possibility is that a2 , b2 , c2 are zero mod 4. Then a, b, c
2
2
2
are even; let A = a2 , B = B2 , c = 2c . Then A2 + B 2 + C 2 = a +b4 +c = n.
3b. Show that if n has the form 4t (8k + 7) for some nonnegative integer t
and integer k, then n cannot be written as the sum of three squares. (In
fact, these are the only numbers that cannot be written as the sum of three
squares, but this is much harder.)
1
2
M3P14 EXAMPLE SHEET 3 SOLUTIONS
Using 3a repeatedly, it suffices to show that no number of the form 8k + 7
can be written as the sum of three squares. Reducing mod 8, we see that
the squares mod 8 are 0, 1, and 4. Thus for any a, b, c, each of a2 , b2 , c2 is
congruent to 0, 1, or 4 mod 8. Let n be the number congruent to 1 mod 8
and m be the number congruent to 4 mod 8. The possibilities are:
• n = m = 0, a2 + b2 + c2 is zero mod 8.
• n = 0, m = 1, a2 + b2 + c2 is 4 mod 8.
• n = 1, m = 0, a2 + b2 + c2 is 1 mod 8.
• n = 1, m = 1, a2 + b2 + c2 is 5 mod 8.
• n = 1, m = 2, a2 + b2 + c2 is 1 mod 8.
• n = 2, m = 0, a2 + b2 + c2 is 2 mod 8.
• n = 2, m = 1, a2 + b2 + c2 is 6 mod 8.
• n = 3, m = 0, a2 + b2 + c2 is 3 mod 8.
Since a2 + b2 + c2 is never 7 mod 8 the claim follows.
4. Prove Wilson’s theorem: If p is prime, then (p − 1)! ≡ −1 (mod p). [Hint:
when multiplying together all the nonzero congruence classes mod p, almost
every class cancels with its inverse. Which ones don’t?]
By definition, (p − 1)! is the product of the positive integers less than p.
For each a ∈ Z with 1 ≤ a ≤ (p − 1), there is a unique multiplicative inverse
of a mod p, and thus a unique a0 with 1 ≤ a0 ≤ (p − 1) and aa0 ≡ 1 (mod
p). Note that (a0 )0 = a.
Moreover, if a0 = a, then a2 ≡ 1 (mod p), so a ≡ ±1 mod p. Thus if a is
between 1 and p − 1 and a0 = a, then a is 1 or (p − 1).
Now we can pair each integer a between 2 and p − 2 with the integer a0 ;
this breaks the integers between 2 and p − 2 into pairs whose product is 1
mod p. Thus the product of the integers between 2 and p − 2 is 1 mod p;
i.e. (p − 2)! ≡ 1 (mod p).
Then (p − 1)! = (p − 1)(p − 2)! ≡ p − 1 ≡ −1 (mod p).
5. Use Fermat descent, starting with 5572 + 552 = 26 · 12049 to write the
prime 12049 as the sum of two squares.
Let a0 = 557, b0 = 55, r0 = 26. In Fermat descent, given ai , bi , ri with
a2i +b2i = ri 12049, we choose ui , vi congruent to ai , bi modulo ri , and between
− r2i and r2i , and set:
ai ui + bi vi
ai+1 =
ri
ai vi − bi ui
bi+1 =
.
ri
Then a2i+1 + b2i+1 = ri+1 12049 for some integer ri+1 less than ri .
In this case we have:
u0 = 11, v0 = 3
a1 = 242, b1 = 41
a21 + b21 = 5 · 12049
M3P14 EXAMPLE SHEET 3 SOLUTIONS
3
r1 = 5
u1 = 2, vi = 1
a2 = 105, b2 = 32
a22 + b22 = 12049
6. For each of the following n, either write n as the sum of two squares, or
prove that it is not possible to do so: 1865, 77077, 609, and 7501.
We have 1865 = 373 · 5. Both these factors are prime, congruent to
1 mod 4, so 1865 is the sum of two squares. We have 5 = (22 + 12 ) =
(2 + i)(2 − i). We also have 373 = 182 + 72 = (18 + 7i)(18 − 7i). Thus
1865 − (2 + i)(18 + 7i)(2 − i)(18 − 7i) = (29 + 32i)(29 − 32i) = 292 + 322 .
We have 77077 = 72 · 112 · 13. We have 13 = 22 + 32 , so 77077 =
(77 · 2)2 + (77 · 3)2 .
We have 609 = 203·3; since 203 is not divisible by 3, 3 divides 609 exactly
once. Since 3 is not a sum of squares, 609 cannot be either.
We have 7501 = 13 · 577. Now 13 = (2 + 3i)(2 − 3i), and 577 = 242 + 12 =
(24 + i)(24 − i). Thus 7501 = (2 + 3i)(24 + i)(2 − 3i)(24 − i) = (45 + 74i)(45 −
74i), so 7501 = 452 + 742 .
√
7. Let ζ = −1+2 −3 , and let Z[ζ] be the subset of C consisting of all complex
numbers of the form a + bζ with a, b integers.
7a. Show that Z[ζ] is closed under addition and multiplication.
The number ζ is a quadratic algebraic integer, as it is a root of x2 +x+1 =
0. This claim thus follows from our result in lecture that shows that Z[α] is
a subring of C whenever α is a quadratic algebraic integer.
7b. Let N : Z[ζ] → C be defined by N (z) = zz. Show that if z ∈ Z[ζ], then
N (z) is an integer.
N (a + bζ) = (a + bζ)(a + bζ) = a2 + ab(ζ + ζ) + b2 ζζ = a2 − ab + b2 .
7c. Show that for any a, b ∈ Z[ζ], with b 6= 0, there exist q, r in Z[ζ] such
that a = qb + r, and either r = 0 or N (r) < N (b).
Let α = ab ; then α is a complex number, and can be written as x + yζ
with x and y real. Let m and n be the nearest integers to x and y, and set
q = m + nζ. Then |x − m|, |y − n| ≤ 12 . We have
a
3
N (q − ) = (x − m)2 + (x − m)(y − n) + (y − n)2 ≤ .
b
4
Thus, setting r = a − bq, we have
a
a
3N (b)
N (r) = N (b(q − )) = N (b)N (q − ) ≤
b
b
4
as desired.
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M3P14 EXAMPLE SHEET 3 SOLUTIONS
7d. Conclude that for any a, b in Z[ζ], a greatest common divisor of a and
b exists.
Part 7c shows that Z[ζ] is a Euclidean domain, so this follows from results
in lecture.
7e. Show that an integer prime p remains prime in Z[ζ] if, and only if, p ≡ 2
(mod 3.) [HINT: show that if p ≡ 1 (mod 3), then there exists n ∈ Z such
that p divides n2 + n + 1.
First suppose that p ≡ 2 mod 3, and that p = uv in Z[ζ]. Then either
N (u) = 1 (u is a unit), N (v) = 1 (v is a unit), or N (u) = N (v) = p. Writing
u = a + bζ, we see that p = a2 + ab + b2 . Testing all possible congruence
classes of a, b mod 3, however, we find that a2 + ab + b2 is always zero or
one mod 3. Thus N (u) cannot equal p and p is irreducible. Since Z[ζ] is a
Euclidean domain, p is prime.
If p = 3, we have 3 = (1 − ζ)(1 − ζ) in Z[ζ], so 3 does not remain prime.
If p ≡ 1 (mod 3), consider the quadratic equation x2 + x + 1 ≡ 0 (mod
p). The discriminant is −3, so this equation has roots if and only if −3 is a
square mod p. Quadratic reciprocity shows that this occurs precisely when
p ≡ 1 mod 3.
So if p ≡ 1 (mod 3), then there exists an n such that n2 + n + 1 ≡ 0 (mod
p); then p divides n2 + n + 1. This means that in Z[ζ], p divides the product
(n − ζ)(n − ζ). Since neither factor is divisible by p, p cannot be prime in
Z[ζ].