18.781 Exam 2 Solutions
Wednesday, November 9, 2005
1. (30 points) Remember that the map for the quadratic form Ax2 +
Bxy + Cy 2 starts out
A+C −B
A
C
A+C +B
a) Conway’s river for the quadratic form x2 − 15y 2 starts out
−14
1
−15
−14
Write similar diagrams for the next six segments of the river.
The next segment of the river corresponds to one of the two forks on the right of
the diagram above. The one that’s on the river is the upper one, which looks like
−15
1
−14
The rule from the diagram with the letters above says that the entry on the right is
determined as the next term in the arithmetic progression beginning −15, 1+(−14);
that is, it’s -11. Continuing in this way, we calculate six segments of the river:
1
−15
−14
−6
1
−11
−11
1
−6
1
−11
−14
1
−11
1
−6
1
−14
−11
−6
−11
1
−14
−15
b) Find a solution of the Diophantine equation x2 − 15y 2 = −14 other
than (x, y) = (±1, ±1).
The numbers on the map are values of the quadratic form on certain vectors
that you can compute as you move along. I have written some of those vectors
by hand in the diagrams above. You need to find a place (not at the beginning of
the map) where the value −14 appears. One place is on the right edge of the fifth
segment above, labelled (11, 3). So a solution is x = 11, y = 3; and indeed one
checks x2 − 15y 2 = 121 − 15 · 9 = 121 − 135 = −14.
2. (30 points) This problem is about the ring R = Z[i] of Gaussian
integers. You may assume things proved in class (like what the primes
in R look like).
a) Find prime factorizations in R of 13 and 17.
The primes in R are ordinary primes congruent to 3 modulo 4, and non-trivial
factors in R of other ordinary primes. The non-trivial factorizations
13 = (3 + 2i)(3 − 2i),
17 = (4 + i)(4 − i)
must therefore be into prime factors. (“Non-trivial” means that neither factor is a
unit; that is, a Gaussian integer of norm 1, which is to say ±1 or ±i.)
b) Suppose that p is a prime congruent to 1 modulo 4. Prove that
the Diophantine equation x2 + y 2 = p has exactly one solution satisfying
x > y > 0.
Any solution of the equation gives a factorization
p = (x + iy)(x − iy),
which must be a prime factorization by the remarks in (a). So if (x0 , y 0 ) is another
solution, then uniqueness of prime factorization implies that
x + iy = ±(x0 ± iy 0 )
or
x + iy = (±i)(x0 ± iy 0 ).
In the first possibility, the hypothesis that x, x0 , y, and y 0 are all positive implies
that the signs are all plus, so x = x0 and y = y 0 . In the second possibility, we get
x2 = (y 0 )2 ,
y 2 = (x0 )2 ,
which contradicts the assumption that x > y > 0 and x0 > y 0 > 0. Therefore there
is at most one solution.
To see that there is a solution, use the fact from class that p is not a Gaussian
prime to write a non-trivial factorization
p = (x + iy)(u + iv).
Taking norms of both sides gives
p2 = (x2 + y 2 )(u2 + v 2 ).
Since the factorization is non-trivial, the factors on the left are not ±1; since they
are positive, they must both be p. So p = x2 + y 2 . Perhaps after changing the signs
of x and y, we may assume that both are non-negative. Since p is not a square,
both are x and y are strictly positive. Since p is odd, x and y must be distinct;
perhaps after interchanging them, we get x > y > 0.
c) Find two solutions of the Diophantine equation x2 + y 2 = 221 satisfying x > y > 0.
Use the factorization 221 = 13 · 17 and the factorizations of 13 and 17 from (a)
to find
(3 + 2i)(4 + i) = 10 + 11i,
(3 + 2i)(4 − i) = 14 + 5i
as two Gaussian integers of norm 221. So the solutions are (11, 10) and (14, 5).
(They’re the only solutions.)
3. (30 points) Suppose that p is an odd prime number, and that
√p is not
prime in the ring of Eisenstein integers S = Z[ζ], with ζ = (−1 + −3)/2.
a) Prove that at least one of the two Diophantine equations
x2 + 3y 2 = p,
or
(2u + 1)2 + 3(2v + 1)2 = 4p
has a solution.
Since p is not an Eisenstein prime, it has a factorization p = αβ with α and β
Eisenstein integers that are not units. Recall that the norm of an Eisenstein integer
is
√
√
√
Norm(x + y −3) = (x + y −3)(x − y −3) = x2 + 3y 2 .
Taking norms in the factorization of p gives p2 = Norm(α) Norm(β). The factors
on the right are non-negative, and they are
√ not 1 (since α and β are not units); so
they must both be p. Writing α = x + y −3, we get
p = Norm(α) = x2 + 3y 2 .
If x and y are integers, we are done. If not, then they are both odd integers divided
by 2: x = (2u + 1)/2, y = (2v + 1)/2, with u and v integers. In this case the
equation p = x2 + 3y 2 becomes
4p = (2u + 1)2 + 3(2v + 1)2 .
b) Prove that in fact both of the equations in (a) have solutions.
Since we know that at least one of the two equations has a solution, it is enough
to construct a solution of the second equation in (a) from a solution of the first,
√and
vice versa. So suppose (x, y) is a solution of the first; that is, (writing α = x+y −3)
that Norm(α) = p. Since ζ is a unit (or by direct computation), it has norm 1; so
Norm(ζα) = p. Now we compute
ζα = (−1 +
√
√
√
−3)(x + y −3)/2 = ((−x + 3y) + (x − y) −3)/2.
Since x2 + 3y 2 is equal to the odd prime p, x and y must have opposite parity.
Therefore the integers 2u + 1 = −x + 3y, 2v + 1 = x − y must both be odd; and
the equation Norm(ζα) = p is just (2u + 1)2 + 3(2v + 1)2 = 4p, as we wished to
show. (The Eisenstein integers help to invent the formulas for u and v in terms of
x and y, but to check that they work is easy algebra.) To proceed in the opposite
direction (from a solution of the second equation to a solution of the first), we can
give a similar argument, or just solve the equations for x and y: x = u + 3v + 2,
y = u+v +1. Again it’s easy algebra to see that these formulas pass from a solution
of the second equation in (a) to a solution of the first.
4. (10 points) This problem is about the ring of Hurwitz integers
{a + bi + cj + dk | 2a, 2b, 2c, 2d ∈ Z
all even or all odd}.
Recall that Norm(a + bi + cj + dk) = a2 + b2 + c2 + d2 , a non-negative integer.
Recall that if α and β are Hurmitz integers with β 6= 0, then there are
Hurwitz integers µ and ρ with
α = µβ + ρ,
Norm(ρ) ≤ Norm(β)/2.
Suppose α = 6 and β = 1 + i + j + k. Find µ and ρ satisfying the equation
above.
As I explained in class, the idea is to pick the Hurwitz integer µ closest to
αβ −1 = (αβ)/ Norm(β) = (6 − 6i − 6j − 6k)/4 = 3/2 − 3i/2 − 3j/2 − 3k/2.
This is a Hurwitz integer, so
µ = 3/2 − 3i/2 − 3j/2 − 3k/2,
ρ = 0.