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18.781 Homework 9
Due: 15th April 2014
Q 1 (3.4(9)). Let f (x, y) = ax2 + bxy + cy 2 be a quadratic form with integral coefficients whose discriminant
d is a perfect square, possibly zero. Show that there are integers h1 , h2 , k1 , k2 such that f (x, y) = (h1 x +
k1 y)(h2 x + k2 y).
Proof. If a = 0, then f (x, y) = y(bx + cy) is a factorization into linear polynomials with integer coefficients
(h1 = 0, k1 = 1, h2 = b, k2 = c). So now assume a 6= 0. Consider the polynomial g(z) = z 2 + ab z + ac ∈ Q[z].
By the assumption that d := b2 − 4ac is a perfect square, say d = k 2 for some k ∈ Z, we see by the quadratic
−b−k
formula that this equation factors into two linear factors in Q[z] (g(z) = (z − −b+k
2a )(z − 2a )). Now,
f (x, y) = ay 2 g(x/y) =
(ax −
−b+k
2 y)(ax
−
−b−k
2 y)
a
As b + k = b − k + 2k, we see that b − k ≡ 0(mod 2) if and only if b + k ≡ 0(mod 2). (b − k)(b + k) = b2 − k 2 =
4ac ≡ 0(mod 2), so 2 | b − k and 2 | b + k. Let −b−k
= l. We then have f (x, y) = (ax−ly)(ax−(l+k)y)
. This
2
a
tells us that l(l + k) = ac. Let r = (a, l), and let a = rs. s | l(l + k) as a | l(l + k) and (s, l) = 1, so s | l + k.
So we see that can write a = rs for some integers r and s such that rt = l and su = l + k for some other
integers t and u, so
f (x, y) = (sx − rl y)(rx − l+k
s )
Q 2 (3.4(10)). Let f (x, y) = ax2 + bxy + cy 2 be a quadratic form with integral coefficients. Show that there
exist integers x0 , y0 , not both 0, such that f (x0 , y0 ) = 0, if and only if the discriminant d of f (x, y) is a
perfect square, possibly 0.
Proof. If f = 0, then we can pick any pair of integers x0 and y0 and the discriminant of f is a square. So
henceforth assume f 6= 0. If the discriminant is a perfect square, then by the previous problem, f (x, y) =
(h1 x + k1 y)(h2 x + k2 y) for some integers h1 , h2 , k1 , k2 , so x0 = k1 , y0 = −h1 satisfies f (x0 , y0 ) = 0 (If f 6= 0,
then h1 and k1 cannot both be zero). Suppose there exist integers x0 , y0 , not both 0, such that f (x0 , y0 ) = 0.
Assume y0 6= 0 (a similar argument works when x0 6= 0). Then we have the equality a xy00 2 + b xy00 + c = 0
over the rational numbers. Multiplying by 4a and completing squares, we see (2a xy00 + b)2 = b2 − 4ac. So the
discriminant d equals l2 /m2 for some integers l and m. l2 /m2 is an integer (i.e. m2 | l2 ), so we must have
m | l. (If m - l, there exists a prime p such that pα | m and pα - l, then p2α | m2 and therefore p2α | l2 but
pα - l implies p2α - l2 , a contradiction).
Q 3 (3.5(1)). Find a reduced form that is equivalent to the form 7x2 + 25xy + 23y 2 .
1
0
−2
1
,
Proof. The discriminant of f is 625 − 644 = −19 which is not a perfect square. Using the matrix
0 1
2
2
we see that the given form is equivalent to 7x − 3xy + y . Using the matrix
, this form is
−1 0
1 −1
equivalent to x2 + 3xy + 7y 2 . Using
, this form is now equivalent to x2 + xy + 5y 2 . This is a
0 1
reduced form.
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Q 4 (3.5(8)). Let f (x, y) = 44x2 − 97xy + 35y 2 . Show that f is equivalent to the form g(x, y) = x(47x − 57y).
Show that n is represented by f if and only if n can be written in the form n = ab where b ≡ 47a(mod 57).
Find the least positive integer n represented by f .
Proof. We see that f (x, y) is equivalent to the form f (x, x + y) = −18x2 − 27xy + 35y 2 = f 0 (x, y) which in
turn is equivalent to f 0 (x + y, y) = −18x2 − 63xy − 10y 2 which in turn is equivalent to −10x2 + 63xy − 18y 2 .
This is in turn equivalent to −10(x + 6y)2 + 63(x + 6y)y − 18y 2 = −10x2 − 57xy, and this in turn equivalent
to −10x2 − 57x(−x + y) = 47x2 − 57xy. n is represented by f if and only if n is represented by g. If n is
represented by g, then n = a(47a − 57y) for some integers a and y. Let b = 47a − 57y, then b ≡ 47a(mod 57).
Conversely, if n = ab and b ≡ 47a(mod 57), then let y ∈ Z be such that b−47a = −57y, then n = a(47a−57y)
which shows n is represented by g and hence by f .
Q 5 (3.7(3)). Show that any positive definite binary quadratic form of discriminant −3 is equivalent to
2
f (x, y) = x2 +
Qxyβ + y . Show that a positive integer n is properly represented by f if and only if n is of the
α
form n = 3
p , where α = 0 or 1 and all the primes p are of the form 3k + 1.
Proof. By Theorem 3.19, if f = ax2 + bxy + cy 2 is a reduced positive definite binary quadratic form of
discriminant −3, then a = 1. As f is reduced, either b = 0 or b = 1. But if b = 0, then the discriminant
is −4c which cannot be −3, so b = 1. b2 − 4ac = −3 now tells us that 1 − 4c = −3, so c = 1. The set of
integers properly represented by a form is the same for equivalent forms. As f is the unique positive definite
reduced quadratic form of discriminant −3, if at all a binary quadratic form of discriminant −3 (which has
to be either positive definite or negative definite) represents a positive integer n, it is represented by f . So
by Theorem 3.13, we see that this is possible if and only if the congruence x2 ≡ −3(mod 4n) is solvable.
x2 ≡ −3(mod 8) is not solvable (as we can see by explcitly writing out square classes modulo 8, or applying
Corollary 2.44), so 2 - n. So the congruence x2 ≡ −3(mod 4n) is solvable if and only if x2 ≡ −3(mod n) is
solvable (x2 ≡ −3(mod 4) is solvable and (4, n) = 1). Also, x2 ≡ −3(mod
Q 9)i is not solvable (again we can
see this explcitly by looking at residue classes modulo 9). Let n = 3α pα
i for some primes pi ≥ 5. By
i
what we have just remarked α ∈ {0, 1}. By Hensel’s lemma, we now see that x2 ≡ −3(mod pα
i ) is solvable
2
i
if and only if x ≡ −3(mod pi ) is solvable for pi ≥ 5 (any solution to this lifts uniquely modulo pα
i as any
solution has to be a unit (as −3 is a unit), and 2 is also a unit, so the derivative 2x evaluated at this solution
is non-vanishing).
p −1
3
−3
i
=
= (−1)(pi −1)/2
(−1)(pi −1)/2 = pi (mod 3)
pi
pi
pi
3
So we see that x2 ≡ −3(mod pi ) is solvable if and only if pi ≡ 1(mod 3)
Q for pi ≥ 5. Putting all of this
together, we see that x2 ≡ −3(mod 4n) is sovable if and only if n = 3α pβ , where α = 0 or 1 and all
the primes p are of the form 3k + 1, and therefore n is properly represented by f if and only if n is of this
particular form.
Q 6 (3.7(6)). Show that any positive definite quadratic form of discriminant −23 is equivalent to exactly
2
2
2
2
2
2
one
of the
form f0 (x, y) = x + xy + 6y , f1 (x, y) = 2x + xy + 3y or f2 (x, y)= 2x − xy + 3y . Show that
if −23
= −1 then p is not represented by any of these forms. Show that if −23
= 1 then p has a total
p
p
of 4 representations by these forms. Show that in the latter case either p has four representations by f0 or
2 representations apiece by f1 and f2 . Determine which of these cases applies when p = 139.
Proof. Every positive definite quadratic form is equivalent to a unique reduced positive definite form, so it
suffices to show that f0 , f1 and f2 are all the distinct reduced positive definite binary quadratic forms. If
f (x, y) =pax2 + bxy + cy 2 is a reduced positive definite binary quadratic form of discriminant −23, then
0 < a < 23/3 by Theorem 3.19. This implies a ∈ {1, 2}. If a = 1, then b ∈ {0, 1}, but b = 0 is not possible
as in this case, the discriminant equals −4ac which cannot equal −23 6≡ 0(mod 4). So if a = 1, then b = 1,
and as b2 − 4ac = −23, c = 6, so we get the reduced form f0 .
If a = 2, then b ∈ {−1, 0, 1, 2}. b 6= 0(mod 2) as b ≡ 0(mod 2) implies the discriminant b2 − 4ac is zero
modulo 2 but −23 is not. So b ∈ {−1, 1}. In this case b2 − 4ac = −23 forces c = 3. So the two reduced
positive definite binary quadratic forms with a = 2 are f1 and f2 .
2
By Corollary 3.14, if
−23
p
= −1, then p is not represented by any form of discriminant −23. If p = 2,
then f0 does not represent it (any non-zero value different from 1 is strictly bigger than 2). If f1 (x, y) = 2,
1
then y = 0 (f1 (x, y) = 2(x + 41 y)2 + (3 − 16
)y 2 ) and x = 1 or x
= −1. Similarly f2 (x,
y) = 2 implies y = 0
and x = 1 or x = −1. So from now on assume p ≥ 3 and −23
= 1. p 6= 23 as −23
= 1.
p
p
If p is represented by a quadratic form, it has to be properlyrepresented
as it is prime. By Theorem 3.26,
=
1
for
each i ∈ {0, 1, 2}. By Theorem
w(fi ) = 2 for all i. We have to calculate Hfi (p) for p such that −23
p
2.20, the number of solutions to u2 ≡ d(mod 4p) is the product of the number of solutions to u2 ≡ d(mod 4)
2
and the number
ofsolutions to u ≡ d(mod p) as (4, p) = 1. Each of these latter equations has 2 solutions
as p > 2 and
−23
p
= 1. So there are four solutions to u2 ≡ d(mod 4p) and therefore there are two integers
h in the range 0 ≤ h < 2p such that h2 ≡ d(mod 4p). The equation rf0 (p) + rf1 (p) + rf2 (p) = wNd (p) = 4
tells us that p has a total of 4 representations by these forms. For each fi , if Hfi (p) 6= 0, then rfi (p) ≥ 2.
f1 (x, y) = f2 (x, −y). So f1 and f2 represent the same integers. So we see that if rf1 (p) 6= 0 if and only
if rf2 (p) 6= 0. If both of these are non-zero, then they both have to equal 2 by our earlier remark, and
rf0 (p) = 0. If they are both zero, then rf0 (p) = 4 because the sum of all three has to equal 4.
332 + 23 = 2 ∗ (4 ∗ 139). Consider the form f (x, y) = 139x2 + 33xy + 2y 2 . This is equivalent to the form
0
f (x, y) = 2x2 − 33xy + 139y 2 . This is equivalent to the form f 0 (x + 8y, y) = 2x2 − xy + 3y 2 , so by what we
have just showed there are two representations of 139 by f1 and f2 each.
Q 7 (5.1(2)). Find all solutions to 10x − 7y = 17.
Proof. (1, −1) is a solution. The general solution is therefore given by (1 + 7x, −1 + 10x) for x ∈ Z by
Theorem 5.1 as (10, −7) = 1.
Q 8 (5.1(10)). Prove that ax + by = c is solvable if and only if (a, b) = (a, b, c) (assume a, b, c are all
non-zero).
Proof. By Theorem 5.1, we see that the equation ax + by = c is solvable if and only if (a, b) | c. If
(a, b, c) = (a, b), then (a, b) | c and the equation is solvable. Now assume the equation is solvable. Then
(a, b) | c. This tells us that (a, b) is a common divisor of a, b and c, so (a, b) | (a, b, c). Any common divisor
of a, b and c is also a common divisor of a and b and as (a, b, c) is one such, (a, b, c) | (a, b). Therefore if the
equation is solvable, (a, b, c) = (a, b).
Q 9 (5.2(1)). Find all solutions in integers of the system of equations
x1 + x2
+4x3 + 2x4
=5
−3x1 − x2
− 6x4
=3
−x1 − x2
+2x3 − 2x4
=1
.
Proof. Adding equations 1 and 3 tells us that 6x3 = 6 or equivalently x3 = 1. Substituting x3 = 1 in
the other two equations now gives us the equivalent system of equations (a) x1 + x2 + 2x4 = 1 and (b)
−3x1 − x2 − 6x4 = 3. Adding 3(a) and (b) gives the equation 2x2 = 6 or x2 = 3. Substituting x2 = 3 in
these two equations now gives us x1 + 2x4 = −2 and −3x1 − 6x4 = 6. Now the second equation is just a
multiple of the first equation, so our original system of equations is equivalent to the system of equations
x1 + 2x4 = −2, x2 = 3 and x3 = 1. So the most general solution to the original system of equations is given
by (−2x4 − 2, 3, 1, x4 ) for x4 ∈ Z.
Q 10 (5.2(2)). For what integers a, b and c does the system of equations
x1 + 2x2
+3x3 + 4x4
=a
x1 + 4x2
+9x3 + 16x4
=b
x1 + 8x2
+27x3 + 64x4
=c
have a solution in integers? What are the solutions if a = b = c = 1?
3
Proof. Replacing equation 2 by equation 2 - equation 1 and replacing equation 3 by equation 3-eqation 1 we
get the equivalent set of equations (a): x1 + 2x2 + 3x3 + 4x4 = a, (b): 2x2 + 6x3 + 12x4 = b − a and (c):
6x2 + 24x3 + 60x4 = c − a. Now replacing equation (c) by equation (c)-3 equation (b) we get the equivalent
set of equations
x1 + 2x2
+3x3 + 4x4
2x2
+6x3 + 12x4
=b−a
=a
6x3 + 24x4
= (c − a) − 3(b − a)
We just applied the row operations as in the book. We will now apply the column operations. First apply
the column operations C1 7→ C1, C2 7→ C2−2C1, C3 7→ C3−3C1, C4 7→ C4−4C1. Now to the new columns,
apply C3 7→ C3 − 3C2, C4 7→ C4 − 6C2. Now to the new columnsapply C4 7→ C4 − 4C3. Applying this se1 −2 −3 + 3(−2) −4 − 6(−2) − 4(−3 + 3(−2))
 0 1
−3
−6 − 4(−3)
quence of operations to the 4×4 identity matrix gives us the matrix 
 0 0
1
−4
0 0
0
1
Let x4 = y4 , x3 = y3 − 4y4 , x2 = y2 − 3y3 + 6y4 , x1 = y1 − 2y2 + 3y3 − 4y4 . Substituting these expressions
into the equations above gives us 6y3 = 2a − 3b + c, 2y2 = b − a and y1 = a. So we see that these equations
are solvable if and only 2 | b − a and 6 | 2a − 3b + c. If a = b = c = 1, then these conditions are satisfied and
we get y3 = 0, y2 = 0, y1 = 1 and y4 is arbitrary. So the general solution to the original set of equations
when a = b = c = 1 is given by (1 − 4y4 , 6y4 , −4y4 , y4 ) for y4 ∈ Z.
Q
11 (5.2(4)).
Let a and b be positive integers, and put g = g.c.d.(a, b), h = l.c.m.(a, b). Show that
a 0
g 0
∼
.
0 b
0 h
Proof. We have to show that we can perform a series of row and column operations that will take the given
matrix with the
matrix
with entries g and h along the diagonal. First add Column 2 to Column1 to get the
s
t
a 0
.
matrix A :=
. Find elements s and t so that as + bt = g. Consider the matrix U =
−b a
b b
g g g bt
.
This matrix has integer entries and has determinant 1 as g | a and g | b. Now U A is the matrix
0 ab
g
bt
Then apply C2 7→ C2 − bt
g C1 ( g is an integer as g | b). After this operation we end up with the matrix
g 0
, and this precisely the matrix we want as gh = ab (Theorem 1.13).
0 ab
g
Q 12 (5.2(5)). Using the preceding problem, or otherwise, show that if D is a diagonal matrix with integral
elements then there is a diagonal matrix S in Smith normal form such that D ∼ S. Deduce that every m × n
matrix A with integral elements is equivalent to a matrix S in Smith normal form.
Proof. Let D = (di δij ) be a n × n diagonal matrix. We will prove that there is a diagonal matrix E = (ei δij )
equivalent to D such that e1 = g.c.d.(e1 , . . . , en ) and ei | ei+1 for all i. This will be by induction on n. Recall
that the g.c.d of two numbers divides the l.c.m. of the same two numbers.
We will first prove that we can perform a sequence of row and column operations to get a diagonal matrix
D̃ whose entry in the (1, 1) is the g.c.d. of all the diagonal entries. This will again be via induction. To
start the induction, notice d1 | d1 . Now assume we have d1 | dj for all j ≤ i and d1 = g.c.d.(d1 , d2 , . . . , di ).
Using row and column operations involving the first row, first column, i + 1st row and i + 1st column
alone, we see that we can get an equivalent diagonal matrix D0 = (d0i δij ) where d01 = g.c.d.(d1 , di+1 ) and
d0i+1 = l.c.m.(d1 , di+1 ) (hence d01 | d0i+1 ) and d0j = dj for j ∈
/ 1, i + 1 by following the steps in the previous
problem. (Such row and column operations only change entries in the (1, 1), (1, i+1), (i+1, 1) and (i+1, i+1)
positions as the original matrix is diagonal). So now we have d01 | d0j for all j ≤ i + 1 and like in the previous
problem we have d01 = g.c.d(d1 , di+1 ) = g.c.d.(d1 , d2 , . . . , di+1 ). This shows the induction step. When i = n,
we get a diagonal matrix D̃ equivalent to D such that the entry in the (1, 1) position divides all the other
4
diagonal entries. Furthermore, the entry in the (1, 1) position is actually equal to the g.c.d (d1 , d2 , . . . , dn ).
Let D̃ = (d˜i δij )
Now if F is the diagonal matrix we get by omitting the first row and first column of D̃, then by the
induction hypothesis, there are unimodular n − 1 × n − 1 matrices S and T such that SF T = (gi δij ) is
0
˜ ˜
˜
a diagonal matrix such that
its g1 = g.c.d.(d2 , d3 , . . . , dn ) and gi | gi+1 for 1 ≤i ≤ n −2. Let S be the
1 0
1 0
unimodular n × n matrix
in block diagonal form and similarly T 0 =
. Then S 0 D̃T 0 is
0 S
0 T
a diagonal matrix with (1, 1) entry e1 = d˜1 = g.c.d.(d1 , d2 , . . . , dn ) and (i, i) entry ei is equal to the gi−1 .
So we have ei | ei+1 for 2 ≤ i ≤ n − 1 by the induction hypothesis. Now, e2 = g.c.d.(d˜2 , d˜3 , . . . , d˜n ) and as
d˜1 | d˜i for all i, and e1 = d˜1 , d˜1 | e2 . This finishes the proof that every diagonal matrix is equivalent to a
diagonal matrix in Smith normal form.
Starting with an arbitrary m × n matrix A, we can perform a sequence of row and column operations to
bring it to a m × n diagonal matrix. Let the rank of this matrix be r. If r = 0, then A = 0 and it is already
in Smith normal form. Now assume r > 0. Let D be the diagonal r × r matrix we get from A by forgetting
all other rows and columns except the first r. By what we have just shown, we can perform row and column
operations to bring this matrix to Smith normal form. By performing the same row and column operations
to the first r rows and r columns of A we can get a matrix equivalent to A in Smith normal form (Such row
and column operations are the same operations we would apply to the diagonal r × r matrix as the original
matrix A is diagonal and therefore aij = 0 if either i or j is strictly greater than r).
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