Algebra A Solutions
A1
Suppose a, b, c > 0 are integers such that
abc − bc − ac − ab + a + b + c = 2013.
(A.1)
Find the number of possibilities for the ordered triple (a, b, c).
Solution. Subtracting 1 from both sides and factoring, we obtain
2012 = (a − 1)(b − 1)(c − 1).
(A.2)
Writing a0 = a − 1 and similarly with b0 , c0 , it suffices to count the ordered integer triples
(a0 , b0 , c0 ) such that a0 , b0 , c0 ≥ 0 and a0 b0 c0 = 2012.
Immediately, we require a0 , b0 , c0 > 0. Let τ (n) denote the number of positive divisors of n.
The prime factorization of 2012 is 22 · 5031 , so the number of possible triples is
X
τ (d) = τ (1) + τ (2) + τ (22 ) + τ (503) + τ (2 · 503) + τ (22 · 503)
d|2012
d>0
which simplifies to 1 + 2 + 3 + 2 + 4 + 6 = 18 .
A2
Find the number of pairs (n, C) of positive integers such that C ≤ 100 and n2 + n + C is a
perfect square.
Remark. This problem was inspired by the “New Year’s Problem” of the 2008-2009
Wisconsin Math Talent Search.
Solution. We can write n2 + n + C = (n + m)2 for some integer m ≥ 1. Expanding and
rearranging, C = m2 + (2m − 1)n, so we have reduced the problem to counting pairs (m, n)
of positive integers such that m2 + (2m − 1)n ≤ 100. This is
10
X
m=1
$
100 − m2
2m − 1
%
= 99 + 32 + 18 + 12 + 8 + 5 + 3 + 2 + 1 + 0
which simplifes to 180 .
1
(A.3)
A3
Let x1 =
√
10 and y1 =
√
3. For all n ≥ 2, let
√
xn = xn−1 77 + 15yn−1
√
yn = 5xn−1 + yn−1 77
(A.4)
(A.5)
Find x65 + 2x45 − 9x45 y52 − 12x25 y52 + 27x25 y54 + 18y54 − 27y56 .
Solution. Let fn = x2n − 3yn2 . We factor
x6n + 2x4n − 9x4n yn2 − 12x2n yn2 + 27x2n yn4 + 18yn4 − 27yn6 = (x2n − 3yn2 )3 + 2(x2n − 3yn2 )2
= fn3 + 2fn2 .
(A.6)
Moreover, f1 = 1, and by induction, we can prove fn = 2fn−1 for all n ≥ 2. In particular,
f5 = 24 , so the answer to the problem is 212 + 3 · 28 = 4096 + 512 = 4608 .
A4
Suppose a, b are nonzero integers such that two roots of x3 + ax2 + bx + 9a coincide, and
all three roots are integers. Find |ab|.
Solution. Write
x3 + ax2 + bx + 9a = (x − r)2 (x − s)
(A.7)
for some integers r, s. Expanding and equating coefficients, we require

2

 r s
= −9a
+ 2rs = b

 2r + s
= −a
r2
(A.8)
Combining the first and third constraints gives r2 s − 18r − 9s = 0. Solving this quadratic
in r, we find that 36(s2 + 9) must be a perfect square, because r, s are integers. Therefore,
s2 + 9 = d2 for some d > 0, whence (d + s)(d − s) = 9. We cannot have s = 0 because this
would imply a = 0 by the first constraint in the display, so the only solutions for s are ±4.
Then (r, s) = ±(6, 4), so |ab| = |(2r + s)(r2 + 2rs)| = 1344 in both cases.
2
A5
Suppose w, x, y, z satisfy
w + x + y + z = 25
wx + wy + wz + xy + xz + yz = 2y + 2z + 193
(A.9)
(A.10)
The largest possible value of w can be expressed in lowest terms as w1 /w2 for some integers
w1 , w2 > 0. Find w1 + w2 .
Remark. This problem was inspired by USAMTS Year 21, Round 4, Problem 2.
Solution. We solve the problem in greater generality. Suppose W, X, Y, Z ≥ 0 satisfy the
modified system
W +X +Y +Z =A
(A.11)
W X + W Y + W Z + XY + XZ + Y Z = B
(A.12)
Expanding (X − Y )2 + (X − Z)2 + (Y − Z)2 ≥ 0, we obtain (X + Y + Z)2 ≥ XY + XZ + Y Z.
Therefore,
B = W (X + Y + Z) + XY + XZ + Y Z
≤ W (A − W ) +
(A − W )2
.
3
(A.13)
After rearranging, we get the following quadratic inequality in W :
2W 2 − AW + (−A2 + 3B) ≤ 0,
(A.14)
√
whence the largest possible value of W is ≤ (A + 9A2 − 24B)/4. But this value is actually
attained at X = Y = Z, as we can see by solving the system where W + 3X = A and
3W X + 3X 2 = B.
Setting (W, X, Y, Z) = (w − 1, x − 1, y + 1, z + 1), we find that the original problem is
equivalent to the problem in W, X, Y, Z with (A,√
B) = (25, 193 + 25 − 2) = (25, 216). Thus,
the largest possible value of w = W + 1 is (25 + 441)/4 + 1 = 46/4 + 1 = 25/2, giving an
answer of 25 + 2 = 27 .
A6
r
Suppose the function ψ satisfies ψ(1) =
2+
q
real x. Determine the greatest integer less than
3
2+
√
Q100
2 and ψ(3x) + 3ψ(x) = ψ(x)3 for all
n
n=1 ψ(3 ).
3
Solution. In order to compute ψ(33 ), we can choose ψ to be any function that satisfies the
stated conditions. In particular, if α is any real number and ψ(x) = 2 cos(αx), then
ψ(3x) = 2 cos(3αx) = 2(4 cos3 (αx) − 3 cos(αx))
= (2 cos(αx))3 − 3(2 cos(αx))
= ψ(x)3 − 3ψ(x).
By applying the half-angle formula twice to 2 cos(π/4) =
r
2 cos(π/16) =
q
2+
2+
(A.15)
√
√
2, we find
2
(A.16)
so we can take α = π/16 = 2π/32.
Since 38 = (34 )2 ≡ (−15)2 ≡ 1 (mod 32), we find that ψ(3n+8 ) = ψ(3n ) for all n. Now,
4
Y
2 cos(3n π/16) = 4 sin(π/8) cos(π/8) = 2 sin(π/4) =
√
2
(A.17)
n=1
√
√
√
Q
Q
n
12 2 = 4096 2. We
and similarly 8n=5 2 cos(3n π/16) = 2. Therefore, 100
n=1 ψ(3 ) = 2
check that 5792 is the largest integer whose square is ≤ 2 · 40962 .
A7
Evaluate
s
r
q
√
2013 + 276 2027 + 278 2041 + 280 2055 + . . .
(A.18)
Remark. This problem was inspired by a theorem of Ramanujan. He famously posed the
similar problem of evaluating
s
r
q
√
1 + 2 1 + 3 1 + 4 1 + ...
(A.19)
in the Journal of the Indian Mathematical Society, with no response after six months. A
good, albeit somewhat romanticized, place to read about Ramanujan is [?].
Solution. We prove that
a+N +n
s
=
r
(a +
n)2
+ aN + N
q
(a + n)2 + a(N + n) + (N + n) (a + n)2 + . . .
4
(A.20)
for all a ∈ R and N, n ∈ Z≥0 with n | N . The result holds for N = 0 and any a, n, and if it
holds for an arbitrary triple (a, N, n), then squaring both sides and simplifying,
r
N + a + 2n =
q
(a + n)2 + a(N + n) + (N + n) (a + n)2 + . . .
(A.21)
so it holds for (a, N + n, n), whence induction finishes the proof.
Our original problem is the case (a, N, n) = (7, 276, 2). The answer is a + N + n = 285 .
A8
The author thanks Will Sawin for his contribution to this solution.
Let S be the set of permutations of {1, 2, . . . , 6}, and let T be the set of permutations of S
that preserve compositions: i.e., if F ∈ T , then
F (f2 ◦ f1 ) = F (f2 ) ◦ F (f1 ).
(A.22)
for all f1 , f2 ∈ S. Find the number of elements F ∈ T such that if f ∈ S satisfies f (1) = 2
and f (2) = 1, then (F (f ))(1) = 2 and (F (f ))(2) = 1.
Solution. We can, and will, write the elements of S in “cycle” notation: For example,
f = (1, 2, 4)(3, 5) means f (1) = 2, f (2) = 4, f (3) = 5, f (4) = 1, f (5) = 3, f (6) = 6. We
write ι for the identity, or null, permutation. Right-to-left composition of permutations can
be expressed as left-to-right1 composition of cycles: For example, (1, 2)(1, 2) = (1, 2)2 = ι
and (1, 2)(1, 3) = (1, 2, 3).
In particular, every permutation can be decomposed into a composition of transpositions,
not necessarily in a unique way. Moreover, all the transpositions are generated by the 6
transpositions (1, 2), (1, 3), . . . , (1, 6).
We want to first describe T completely. For all f ∈ S, there is a uniquely associated element
Ff ∈ T defined by Ff (g) = f −1 ◦ g ◦ f . However, these do not account for all of T . To get
the others, it is necessary to describe S in more detail:
6!
We can classify the elements of S according to “cycle type.” For example, there are 2·4!
= 15
6!
transpositions, or type-2 cycles; there are 3·3! = 40 type-3 cycles, such as (1, 2, 3); there
6!
are 2·2·2·3!
= 15 type-2, 2, 2 cycles, such as (1, 2)(3, 4)(5, 6); and so on. The main thing to
notice is that there are as many transpositions as type-2, 2, 2 cycles, which suggests that a
suitable bijection between them could extend to an element of T , since the transpositions
generate T .
1
We introduce this change in the direction of composition because it makes the results from cycle
composition more clear.
5
Let S2 be the set of transpositions, and let S2,2,2 be the set of type-2, 2, 2 cycles. By the
associativity of composition of permutations, a bijection S2 → S2,2,2 extends to an element
F0 ∈ T if and only if it preserves compositions on S2 , i.e., F0 (f1 ◦f2 ) = F0 (f1 )◦F0 (f2 ) for all
f1 , f2 ∈ S2 . The composition of two transpositions is either ι, a type-3 cycle, or a type-2, 2
cycle. The composition of two type-2, 2, 2 cycles is either ι; a type-2, 2 cycle, if they share
exactly one transposition; or a type-3, 3 cycle, if they share no transpositions. Therefore, it
suffices to find a bijection S2 → S2,2,2 that sends disjoint transpositions to 2, 2, 2-cycles that
share exactly 1 transposition, and transpositions with 1 number in common to type-2, 2, 2
cycles that share no transpositions, such that in the latter case, the induced map on the
compositions is well-defined: i.e., F0 ((1, 2)(1, 3)) = F0 ((2, 3)(1, 2)) = F ((1, 3)(2, 3)).
These conditions are met if, for example, we let F0 be defined by
(1, 2)
(1, 3)
(1, 4)
(1, 5)
(1, 6)
...
7→
7
→
7→
7→
7→
(1, 2)(3, 6)(4, 5)
(1, 6)(2, 4)(3, 5)
(1, 3)(2, 5)(4, 6)
(1, 5)(2, 6)(3, 4)
(1, 4)(2, 3)(5, 6)
...
(A.23)
(Our choice of the image of (1, 2) follows the choice in [?].) It is useful to note that F0 ◦ F0
is the identity permutation on elements of S.
We can check that a given element of T takes the form Ff for some f ∈ F if and only if it
sends isolated transpositions to isolated transpositions. Therefore, F0 does not arise from
an f in this way. However, we claim that
T = {Ff : f ∈ S} ∪ {Ff ◦ F0 : f ∈ S}.
(A.24)
First, observe that if C is the set of all elements of S of a given cycle type, then any element
of T , being invertible, must map C bijectively onto another set of the same form, i.e.,
consisting of all elements of another given cycle type. Such a set is called a conjugacy class
of elements of S. But one verifies that the only conjugacy classes of size 15 in S are S2 and
S2,2,2 . Altogether, if we take an arbitrary G ∈ T , then either G keeps the transpositions in
place or exchanges them with the type-2, 2, 2 cycles; in the former case, G = Ff for some
f , and in the latter case, G ◦ F0 keeps transpositions in place, so G ◦ F0 = Ff for some f ,
whence G = G ◦ F0 ◦ F0 = Ff ◦ F0 . This proves the claim.
Finally, we can compute the answer to the problem. Let S 0 be the set of g ∈ S such that
g(1) = 2 and g(2) = 1.
If Ff maps S 0 onto itself, then by computation, any cycle notation for f must either contain
(1, 2), or else cannot contain any cycles involving either 1 or 2. There is a bijection between
the possibilities for f in the former case and those in the latter case; moreover, in the latter
6
case, counting by cycle type shows there are 1 + 6 + 8 + 3 + 6 = 24 possibilities, so that
in total, there are 48 possibilities. Conversely, any such f yields an Ff that maps S 0 onto
itself.
By the example above, we know we can choose F0 such that F0 ((1, 2)) = (1, 2)(3, 6)(4, 5).
Thus, any element Ff ◦ F0 such that Ff sends any one of (1, 2), (3, 6), or (4, 5) to (1, 2) will
map S 0 onto itself. So there are also precisely 3(48) = 144 elements of the form Ff ◦ F0
that map S 0 onto itself. Altogether the answer to the problem is 48 + 144 = 192 .
Remark. For further information about this solution, we advise the reader to consult
[?]. For more about the mathematical objects at work in the problem, we recommend the
beautiful exposition in Ch. 1-3 of [?]. In particular, S and T have the structure of groups;
S is S6 , the symmetric group on 6 letters, and T = Aut(S6 ), its automorphism group.
References
[He]
I. N. Herstein. Abstract Algebra. 3rd Ed., Prentice Hall (1996).
[JR]
G. Janusz, J. Rotman. “Outer Automorphisms of S6 .” Amer. Math. Mon., Vol. 89,
No. 6 (1982), 407-410.
[Ka]
R. Kanigel. The Man Who Knew Infinity: a Life of the Genius Ramanujan. New
York: Charles Scribner’s Sons (1991). ISBN 0-684-19259-4.
7