A1 A grasshopper starts at the origin in the coordinate plane and makes a sequence of hops. Each
hop has length 5, and after each hop the grasshopper is at a point whose coordinates are both
integers; thus, there are 12 possible locations for the grasshopper after the first hop. What is
the smallest number of hops needed for the grasshopper to reach the point (2021, 2021)?
Answer: 578.
Solution: Each hop can be described by a displacement vector hp, qi with p2 + q 2 = 25 ; the
twelve possible vectors are
h3, 4i; h−3, 4i; h3, −4i; h−3, −4i; h4, 3i; h−4, 3i; h4, −3i; h−4, −3i; h5, 0i; h−5, 0i; h0, 5i; h0, −5i.
One way to write the total displacement as a sum of 578 of these vectors is
h2021, 2021i = 288 · h3, 4i + 288 · h4, 3i + h0, 5i + h5, 0i.
To show that it cannot be done with fewer, note that each hop can increase the sum of the
grasshopper’s coordinates by at most 3 + 4 = 7. Because this sum has to reach
2021 + 2021 = 4042 = 7 · (577) + 3 ,
at least 578 hops are needed.
1
A2 For every positive real number x, let
g(x) = lim (x + 1)r+1 − xr+1
r→0
r1
.
g(x)
.
x
Answer: e.
Find lim
x→∞
Solution: Note that for r > −1 and any positive x, we have (x + 1)r+1 − xr+1 > 0. Thus, by
the continuity of the logarithm,
1
log g(x) = lim log ((x + 1)r+1 − xr+1 ) r
r→0
= lim
r→0
1
log (x + 1)r+1 − xr+1 .
r
Applying L’Hôpital’s rule, we get
(x + 1)r+1 log(x + 1) − xr+1 log x
r→0
(x + 1)r+1 − xr+1
(x + 1) log(x + 1) − x log x
= log (x + 1)x+1 x−x ,
=
(x + 1) − x
x
1
g(x) = (x + 1)x+1 x−x = (x + 1) 1 +
.
x
log g(x) = lim
Finally,
g(x)
x+1
= lim
x→∞ x
x→∞
x
lim
2
1+
1
x
x
= 1 · e = e.
so
A3 Determine all positive integers N for which the sphere
x2 + y 2 + z 2 = N
has an inscribed regular tetrahedron whose vertices have integer coordinates.
Answer: A necessary and sufficient condition is that N be of the form N = 3m2 , where m is
a positive integer.
Solution 1: To see that the condition is sufficient, note that the four points
(−m, −m, −m), (m, m, −m), (m, −m, m), (−m, m, m)
are the vertices of a regular tetrahedron inscribed in the sphere x2 + y 2 + z 2 = 3m2 .
To show that the condition is necessary, we will use two lemmas:
Lemma 1. If T is a tetrahedron whose vertices have integer coordinates, then its volume is
of the form V (T ) = D/6 for some integer D.
Lemma 2. The
√ volume of a regular tetrahedron T inscribed in a sphere of radius R is given
8 3R3
by V (T ) =
.
27
Assuming that the sphere x2 + y 2 + z 2 = N has an inscribed regular tetrahedron T whose
√
vertices have integer coordinates, we can combine the results of these lemmas (for R = N )
to get
√
16 3 N 3/2
16N √
3N .
D = 6 V (T ) =
=
9
9
√
Because D is an integer, it follows that 3N is a rational number. Thus the prime factorization
of N must contain an odd number of factors 3 and an even number of factors p for any other
prime p ; therefore, N = 3m2 for some positive integer m.
Proof of Lemma 1: Let P , Q, R, and S be the vertices of the tetrahedron. As these are all
lattice points, the three vectors
P~Q = hq1 , q2 , q3 i,
P~R = hr1 , r2 , r3 i,
P~S = hs1 , s2 , s3 i
have integer components. We can use a triple product to express the volume as

q1
1 ~
1
~
~

V = |P Q · (P R × P S)| =
det r1
6
6
s1
q2
r2
s2

q3
r3  ,
s3
and as the determinant is an integer, we are done.
Proof of Lemma 2: If a regular tetrahedron T is inscribed in a sphere of radius R, we can
choose a coordinate system in which the vertices of T are given by
P = (−d, −d, −d),
Q = (d, d, −d),
R = (d, −d, d),
S = (−d, d, d)
with 3d2 = R2 . We then have
P~Q = h0, 2d, 2di,
P~R = h2d, 0, 2di,
3
P~S = h2d, 2d, 0i,
and the volume is

√
0 2d 2d
8 3
1 ~
1
8 3R3
~
~


=
det
V = |P Q · (P R × P S)| =
d =
.
2d 0 2d
6
6
3
27
2d 2d 0

Solution 2: As in Solution 1, the condition on N is sufficient. To show that it is necessary,
note that if
vi = (xi , yi , zi ), i = 1, 2, 3, 4
are the vertices of a regular tetrahedron inscribed in the sphere x2 + y 2 + z 2 = N , we can add
the four antipodal points
wi = (−xi , −yi , −zi ),
i = 1, 2, 3, 4
to get the eight vertices of a cube. (This is easily seen by choosing an alternate coordinate
system as in the proof of Lemma
p sphere, its
√ 2 above.) Because this cube is inscribed in the
space diagonals have length 2 N ; therefore, each edge of the cube has length 2 N/3 and its
3
p
N/3 . But the volume of the cube is the determinant of three vectors with
volume is 8
integer coordinates, so it is an integer, and as in Solution 1 it follows that N = 3m2 for some
positive integer m.
4
A4 Let
ZZ
I(R) =
1 + y2
1 + 2x2
−
4
2
2
4
1 + x + 6x y + y
2 + x4 + y 4
dx dy .
x2 +y 2 ≤R2
Find
lim I(R) ,
R→∞
or show that this limit does not exist.
√
π 2 log 2
Answer: The limit exists and equals
.
2
Solution: First we symmetrize the integrand. Let
f (x, y) =
x4
1+
f (x, y) + f (y, x) =
1 + 2x2
1 + y2
−
,
2
2
4
+ 6x y + y
2 + x4 + y 4
so that
2 + 2(x2 + y 2 )
2 + x2 + y 2
−
1 + x4 + 6x2 y 2 + y 4
2 + x4 + y 4
and thus
2 + 2(x2 + y 2 )
2 + x2 + y 2
−
dx dy.
1 + x4 + 6x2 y 2 + y 4
2 + x4 + y 4
ZZ
2 I(R) =
x2 +y 2 ≤R2
Now consider the “first part” of this double integral, say
ZZ
2 + 2(x2 + y 2 )
J(R) =
dx dy.
1 + x4 + 6x2 y 2 + y 4
x2 +y 2 ≤R2
Let u = x − y and v = x + y. Then
u2 +v 2 = (x+y)2 +(x−y)2 = 2(x2 +y 2 ),
and
∂(u, v)
1
=
1
∂(x, y)
u4 +v 4 = (x+y)4 +(x−y)4 = 2x4 +2y 4 +12x2 y 2
−1
= 2 , so
1
2 + 2(x2 + y 2 )
2 dx dy
2 + 2x4 + 12x2 y 2 + 2y 4
ZZ
J(R) =
x2 +y 2 ≤R2
2 + u2 + v 2
du dv.
2 + u4 + v 4
ZZ
=
u2 +v 2 ≤2R2
Note that if we rename the variables in this last integral x, y instead of u, v, the integrand
will be the same as the integrand of the “second part” of the double integral for 2 I(R) above.
Thus we can recombine the parts to get
ZZ
2 + x2 + y 2
2 I(R) =
dx dy.
2 + x4 + y 4
R2 <x2 +y 2 ≤2R2
Converting to polar coordinates, we get
Z
2π
Z
√
R 2
2 I(R) =
t=0
r=R
2 + r2
r dr dt.
2 + r4 (cos4 t + sin4 t)
5
As R → ∞, throughout the range of integration r also goes to infinity and
2+
2r + r3
1
=
+ O(1/r3 ) ,
4
4
t + sin t)
r(cos t + sin4 t)
r4 (cos4
where the error term makes a vanishing contribution to the integral. So
# Z
"Z √
2π
R 2
dr
dt
2 I(R) ∼
.
4
4
r
r=R
t=0 cos t + sin t
Now
Z
√
R 2
r=R
√
√
dr
= log(R 2) − log(R) = log( 2) =
r
1
2
log 2
and
Z
2π
t=0
dt
=
cos4 t + sin4 t
Z
2π
t=0
2π
Z
=
t=0
dt
=
1 − 2 sin2 t cos2 t
Z
2π
t=0
2 dt
2 − sin2 (2t)
2 dt
.
2
2 cos (2t) + sin2 (2t)
π
and is also even, so we can proceed as follows:
2
Z π/4
2 dt
2 dt
=
8
2 (2t) + sin2 (2t)
2 cos2 (2t) + sin2 (2t)
2
cos
t=0
The integrand is periodic with period
Z
2π
t=0
Z π/4
dt
=
4
cos4 t + sin4 t
t=−π/4
Z π/4
16 sec2 (2t) dt
=
2 + tan2 (2t)
t=0
Z ∞
Z ∞
8 dw
4 dw
=
=
2
2
w=0 2 + w
w=0 1 + w /2
∞
√
√
√
= 2π 2 .
= 4 2 tan−1 (w/ 2)
w=0
So
√
√
1
lim 2 I(R) = ( log 2)(2π 2) = π 2 log 2 ,
R→∞
2
6
and
√
π 2 log 2
lim I(R) =
.
R→∞
2
A5 Let A be the set of all integers n such that 1 ≤ n ≤ 2021 and gcd(n, 2021) = 1. For every
nonnegative integer j, let
X
S(j) =
nj .
n∈A
Determine all values of j such that S(j) is a multiple of 2021.
Answer: All j that are not multiples of 42 or 46.
Solution: Note that modulo 2021, the set A consists precisely of the elements of the multiplicative group. Multiplying by an element of that group permutes the elements, so if x is
relatively prime to 2021, then
X
X
xj · S(j) =
(xn)j ≡
mj ≡ S(j) (mod 2021).
n∈A
m∈A
Therefore,
(xj − 1) S(j) ≡ 0 (mod 2021).
Also note that 2021 = 2025 − 4 = 452 − 22 = 43 · 47 gives the prime factorization of 2021. Let
x be a primitive root modulo 43 (that is, an integer between 1 and 42 that is a generator of
the cyclic group (Z/43 Z)∗ , which is the multiplicative group of the field with 43 elements).
Then xj − 1 ≡ 0 (mod 43) if and only if j is a multiple of 42 ; also, x is relatively prime to
2021. In particular, if j is not a multiple of 42 we have
(xj − 1) S(j) ≡ 0 (mod 2021) ⇒ (xj − 1) S(j) ≡ 0 (mod 43) ⇒ S(j) ≡ 0 (mod 43) .
Similarly, if y is a primitive root modulo 47 and y 6= 43, we have
(y j − 1) S(j) ≡ 0 (mod 2021) ⇒ (y j − 1) S(j) ≡ 0 (mod 47) ⇒ S(j) ≡ 0 (mod 47)
whenever j is not a multiple of 46. So if j is not a multiple of 42 or 46, then S(j) is a multiple
of both 43 and 47, hence of 2021.
Conversely, suppose that j is a multiple of 42. Then nj ≡ 1 (mod 43) for all n in the sum,
and S(j) is therefore not a multiple of 43 (or of 2021), as
X
S(j) ≡
1 = 42 · 46 · 1 ≡ −3 ≡ 40 (mod 43).
n∈A
Similarly, if j is a multiple of 46, then S(j) ≡ 5 (mod 47).
7
A6 Let P (x) be a polynomial whose coefficients are all either 0 or 1. Suppose that P (x) can be
written as the product of two nonconstant polynomials with integer coefficients. Does it follow
that P (2) is a composite integer?
Solution: Yes, we will show that P (2) must be composite. Let P (x) = F (x)G(x) have
degree N , where F (x) and G(x) are nonconstant polynomials with integer coefficients, and
suppose that P (2) = p were prime. Then either F (2) or G(2) would be a unit, so without
loss of generality we may assume that G(2) = 1. We have N ≥ 2 because F (x) and G(x) are
nonconstant, and we can write
N
X
P (x) =
σn xn ,
n=0
where each σn is either 0 or 1 and σN = 1.
Because the coefficients of P (x) are nonnegative integers, P (x) cannot have a positive real
root, so G(x) cannot have a positive real root either. Thus, as G(2) is positive, G(x) must be
positive for all x > 0. In particular, because σN = 1 is the product of the leading coefficients
of F (x) and G(x), the polynomial G(x) must be monic. Let r1 , . . . , rk be the (complex, not
necessarily distinct) roots of G(x), so that
G(x) =
k
Y
(x − rj ) .
j=1
Consider the integer G(1). Because G(x) > 0 for x > 0, we have G(1) ≥ 1 = G(2). In
particular, |G(1)| ≥ |G(2)|, and using the factorization of G(x) we get
k
Y
|1 − rj | ≥
k
Y
|2 − rj | .
j=1
j=1
It follows that G(x) must have at least one root ρ with |ρ − 1| ≥ |ρ − 2|, which is equivalent
to Re(ρ) ≥ 23 . This implies that |ρ| ≥ 32 ; note that ρ is also a root of P (x).
First consider the case N = 2. Dividing P (ρ) = 0 by ρ yields
ρ + σ1 +
We have Re(ρ) ≥
But then
3
2
σ0
= 0.
ρ
> 0 and hence
ρ̄
1
1
= Re
= 2 Re (ρ) > 0.
Re
2
ρ
|ρ|
|ρ|
σ0
= 0,
Re(ρ) ≤ Re ρ + σ1 +
ρ
which is a contradiction. (Alternatively, one can check the four possible polynomials P (x) of
degree 2 with coefficients from {0, 1}.)
For N > 2, we again divide P (ρ) = 0 by ρN −1 , which now yields
ρ + σN −1 +
σN −2
σN −3
σ0
= − 2 − · · · − N −1 .
ρ
ρ
ρ
Once again, the terms on the left have nonnegative real parts. The triangle inequality gives
σN −2
σN −2
σN −3
σ0
Re(ρ) ≤ Re ρ + σN −1 +
≤ ρ + σN −1 +
≤
+ · · · + N −1 ,
ρ
ρ
|ρ|2
|ρ|
8
and we can estimate the sum on the right using an infinite geometric series:
σ0
1
1
σN −3
+ · · · + N −1 ≤ 2 + · · · + N −1
|ρ|2
|ρ|
|ρ|
|ρ|
1
1
1
≤ 2 1+
+ ···
+
|ρ|
|ρ| |ρ|2
1
1
1
= 2·
=
.
|ρ| 1 − 1/|ρ|
|ρ|(|ρ| − 1)
But
1
3
is a decreasing function of x for x > 1, and |ρ| ≥ Re(ρ) ≥ , so we get
x(x − 1)
2
3
1
≤ Re(ρ) ≤
≤
2
|ρ|(|ρ| − 1)
3 3
2(2
4
1
= ,
3
− 1)
a contradiction.
Remark. There are polynomials like x7 + x2 + x + 1 = (x + 1)(x2 + 1)(x4 − x3 + 1) or
x7 + x3 + x2 + x + 1 (which is irreducible) which have roots with real part greater than 1. The
polynomial x11 + x3 + x2 + x + 1 has a root r with |r − 2| < 1. Hence one needs to take some
care with this argument.
9