2.4 On an Algebraic IdentityThe pol

2.4 On an Algebraic Identity
The polynomial X2+1 is irreducible over R, but, of course, X4+1 is not. To factor it,
complete the square then view the result as a difference of two squares:
X4+1 = (X4+2X2+1)−2X2 = (X2 +1)2−(
√
2X)2
= (X2+
√
2X +1)(X2−
√
2X +1).
From this we can derive the Sophie Germain identity
X4+4Y4 = (X2 +2XY +2Y2)(X2−2XY +2Y2),
2.4. On an Algebraic Identity 49
with an alternative version
X4+
1
4
Y4 =

X2+XY +
1
2
Y2

X2−XY +
1
2
Y2

.
Knowing this identity can be useful in some situations. Here are two examples.
Evaluate the sum
nΣ
k=1
4k
4k4 +1
.
By using the above identity, we obtain
nΣ
k=1
4k
4k4+1
=
nΣ
k=1
(2k2+2k+1)−(2k2−2k+1)
(2k2+2k+1)(2k2−2k+1)
=
nΣ
k=1

1
2k2−2k+1
− 1
2(k+1)2−2(k+1)+1

= 1− 1
2n2+2n+1
,
and we are done.
Given an n × n matrix A with the property that A3 = 0, prove that the matrix
M = 12
A2+A+In is invertible.
Indeed, the inverse of this matrix is 12
A2 −A +In, since the product of the two
gives 14
A4 +In, which is equal, by hypothesis, to the identity matrix. Of course, the
conclusion can also be derived from the fact that the eigenvalues of M are nonzero,
which is a direct consequence of the spectral mapping theorem. Or, we can notice that
In+A+A2/2 = eA, and its inverse is e−A = I−A+A2/2.
We give below more examples that make use of the Sophie Germain identity.
1. Prove that for every integer n > 2, the number 22n−2+1 is not a prime number.
2. Prove that any sequence satisfying the recurrence relation xn+1+xn−1 =
√
2xn is
periodic.
3. Compute the sum
nΣ
k=1
k2 − 12
k4 + 14
.
4. Evaluate
(14+ 14
)(34+ 14
) · · · ((2n−1)4+ 14
)
(24+ 14
)(44 + 14
) · · · ((2n)4+ 14
)
.
5. Show that there are infinitely many positive integers a such that for any n, the
number n4+a is not prime.
50 Chapter 2. Algebra and Analysis
6. Show that n4+4n is prime if and only if n = 1.
7. Consider the polynomial P(X) = X4 + 6X2 − 4X + 1. Prove that P(X4) can
be written as the product of two polynomials with integer coefficients, each of
degree greater than 1.
8. Prove that for any integer n greater than 1, n12+64 can be written as the product
of four distinct positive integers greater than 1.
9. Let m and n be positive integers. Prove that if m is even, then
mΣ
k=0
(−4)kn4(m−k)
is not a prime number.
10. Find the least positive integer n for which the polynomial
P(x) = xn−4+4n
can be written as a product of four non-constant polynomials with integer
coefficients.
2.5 Systems of Equations
For this section, we have selected non-standard systems of equations. The first example
involves just algebraic manipulations.
Prove that the only positive solution of
x+y2+z3 = 3,
y+z2+x3 = 3,
z+x2+y3 = 3
is (x,y, z) = (1,1,1).
From the difference of the first two equations, we obtain that
x(1−x2)+y(y−1)+z2(z−1) = 0.
From the difference of the last two equations, we obtain that
y(1−y2)+z(z−1)+x2(x−1) = 0.
Multiplying this equation by z and subtracting it from the one above yields
x(x−1)(1+x+xz)= y(y−1)(1+z+yz).
Similarly
y(y−1)(1+y+yx)= z(z−1)(1+x+zx).
2.5. Systems of Equations 51
From the last two relations, it follows that if x,y, and z are positive, then x,y, and z
are all equal to 1, all less than 1, or all greater than 1. The last two possibilities are
excluded, for x+y2+z3 = 3, and the result follows.
The second example is from the 1996 British Mathematical Olympiad.
Find all solutions in positive real numbers a,b,c,d to the following system:
a+b+c+d = 12,
abcd = 27+ab+ac+ad+bc+bd+cd.
Using the Arithmetic Mean–Geometric Mean (AM–GM) inequality in the second
equation, we obtain
abcd ≥ 27+6
√
abcd.
Moving everything to the left and factoring the expression, viewed as a quadratic polynomial
in
√
abcd, yields
(
√
abcd+3)(
√
abcd−9) ≥ 0.
This implies
√
abcd ≥ 9, which combined with the first equation of the system
gives
4 √
abcd ≥ a+b+c+d
4
.
The AM–GM inequality implies that a = b = c = d = 3 is the only solution.
And now a problem with a surprising solution from the 1996 Vietnamese Mathematical
Olympiad.
Find all positive real numbers x and y satisfying the system of equations
√
3x

1+
1
x+y

= 2,

7y

1− 1
x+y

= 4
√
2.
It is natural to make the substitution
√
x = u,
√
y = v. The system becomes
u

1+
1
u2+v2

=
√2
3
,
v

1− 1
u2+v2

=
4
√
√ 2
7
.
But u2+v2 is the square of the absolute value of the complex number z = u+iv. This
suggests that we add the second equation multiplied by i to the first one. We obtain
u+iv+
u−iv
u2+v2 =

√2
3
+i
4
√
√ 2
7

.
52 Chapter 2. Algebra and Analysis
The quotient (u−iv)/(u2+v2) is equal to ¯z/|z|2 = ¯z/(z¯z) = 1/z, so the above equation
becomes
z+
1
z
=

√2
3
+i
4
√
√ 2
7

.
Hence z satisfies the quadratic equation
z2 −

√2
3
+i
4
√
√ 2
7

z+1 = 0
with solutions

√1
3
± √2
21

+i

2
√
√ 2
7
±
√
2

,
where the signs + and − correspond.
This shows that the initial system has the solutions
x =

√1
3
± √2
21
2
, y =

2
√
√ 2
7
±
√
2
2
,
where the signs + and − correspond.
The systems below are to be solved in real numbers, unless specified otherwise.
1. Solve the system of equations
x+log(x+

x2+1) = y,
y+log(y+

y2+1) = z,
z+log(z+

z2 +1) = x.
2. Solve the system
log(2xy) = logxlogy,
log(yz) = logylogz,
log(2zx) = logz logx.
3. Solve the system of equations
xy+yz+zx = 12,
xyz = 2+x+y+z
in positive real numbers x,y, z.
2.5. Systems of Equations 53
4. Find all real solutions to the system of equations
4x2
4x2+1
= y,
4y2
4y2+1
= z,
4z2
4z2 +1
= x.
5. Find ax5+by5 if the real numbers a,b,x, and y satisfy the system of equations
ax+by= 3,
ax2+by2 = 7,
ax3+by3 = 16,
ax4+by4 = 42.
6. Find all solutions to the system of equations
6(x−y−1) = 3(y−z−1) = 2(z−x−1) = xyz−(zyz)−1
in nonzero real numbers x,y, z.
7. Find all integer solutions to the system
3 = x+y+z= x3+y3+z3.
8. Solve the system
x+
2
x
= 2y,
y+
2
y
= 2z,
z+
2
z
= 2x.
9. Solve the system of equations
(x+y)3 = z,
(y+z)3 = x,
(z+x)3 = y.
10. Solve the system
x2 −|x| = |yz|,
y2 −|y| = |zx|,
z2 −|z| = |xy|.
54 Chapter 2. Algebra and Analysis
11. Find the solutions to the system of equations
x+y
+{z}= 1.1,
x
+{y}+z= 2.2,
{x}+y+z
= 3.3,
where 
and {} denote respectively the greatest integer and fractional part functions.
12. For a given complex number a, find the complex solutions to the system
(x1 +x2+x3)x4 = a,
(x1 +x2+x4)x3 = a,
(x1 +x3+x4)x2 = a,
(x2 +x3+x4)x1 = a.
13. Find all real numbers a for which there exist nonnegative real numbers x1,x2,x3,x4,x5
satisfying the system
5Σ
k=1
kxk = a,
5Σ
k=1
k3xk = a2,
5Σ
k=1
k5xk = a3.
14. Solve the system of equations
x3−9(y2−3y+3)= 0,
y3−9(z2−3z+3)= 0,
z3 −9(x2−3x+3)= 0.
15. Solve the system
ax+by= (x−y)2,
by+cz= (y−z)2,
cz+ax = (z−x)2,
where a,b,c > 0.
16. Let a,b,c be positive real numbers, not all equal. Find all solutions to the system
of equations
x2 −yz = a,
y2 −zx = b,
z2 −xy = c,
in real numbers x,y, z.
2.6. Periodicity 55
2.6 Periodicity
Periodicity plays an important role inmathematics, and for this reason we have included
some problems involving it. The first example shows how periodicity can be used to
give a short proof of the Hermite identity for the greatest integer function:
x
+

x+
1
n

+· · ·+

x+
n−1
n

= nx
,
for all x ∈ R and n ∈ Z.
The proof proceeds as follows. Define f : R→N,
f (x) = x
+

x+
1
n

+· · ·+

x+
n−1
n

−nx
.
One can easily check that f (x) = 0 for x ∈

0, 1n

. Also
f

x+
1
n

=

x+
1
n

+· · ·+

x+
n−1
n

+

x+
n
n


−nx+1
= f (x).
Hence f is periodic, with period 1n
. This shows that f is identically equal to 0, which
proves the identity.
The second example concerns the computation of a sum of binomial coefficients.
Compute the sum

n
0

−

n−1
1

+

n−2
2

−

n−3
3

+· · ·
Denote our sum by Sn. Then
Sn−Sn−1+Sn−2 =

n
0

−

n−1
1

+

n−2
2

−

n−3
3

+· · ·
−

n−1
0

+

n−2
1

−

n−3
2

+

n−4
3

−· · ·
+

n−2
0

−

n−3
1

+

n−4
2

−

n−5
3

+· · · .
Since

n0

=

n−1
0

, the first terms of Sn and Sn−1 cancel. If we group the kth term of
Sn with the kth term of Sn−1 and the (k − 1)st term of Sk−2, we obtain
(−1)k−1

nk

−

n−1
k

−

n−1
k−1

, which is zero by the recurrence formula for binomial
coefficients. Hence all terms cancel. It follows that Sn −Sn−1 +Sn−2 = 0. Add
Sn−1 −Sn−2 +Sn−3 = 0 to this equality to obtain Sn = −Sn−3. This shows that Sn is
periodic of period 6.
Thus it suffices to compute Sn for n = 1, 2, 3, 4, 5, and 6, the other values repeating
with period 6. We obtain S6n+1 = S1 = 1, S6n+2 = S2 = 0, S6n+3 = S3 = −1, S6n+4 =
S4 = −1, S6n+5 = S5 = 0, and S6n = S6 = 1.
56 Chapter 2. Algebra and Analysis
The third example is a short-listed problem from the International Mathematical
Olympiad.
Let f : R→R be a bounded function such that
f

x+
1
6

+ f

x+
1
7

= f (x)+ f

x+
13
42

.
Show that f is periodic.
Define g : R→R, g(x) = f (x+ 16
)− f (x). The condition from the statement tells
us that g(x+ 17
) = g(x). Therefore g(x+1) = g(x). If we let h : R → R, h(x) =
f (x+1)− f (x) = g(x)+g(x+ 16
)+· · ·+g(x+ 56
), then we also have that h(x+1) =
h(x). Thus we obtain h(x+k) = h(x) for all k ∈ N. Since f (x+k)− f (x) = h(x)+
h(x+1)+· · ·+h(x+k−1), we obtain f (x+k)− f (x) = kh(x). And because f is
bounded, kh(x) is bounded as well for all positive integers k, which is possible only if
h is identically equal to zero. It follows that f (x+1) = f (x) for all x, so f is periodic
with period 1.
Here are more problems involving periodicity.
1. Let f : R→R−{3} be a function with the property that there exists ω > 0 such
that
f (x+ω) =
f (x)−5
f (x)−3
, for all x ∈ R.
Prove that f is periodic.
2. Let f : R→R be a function that satisfies
(a) f (x+y)+ f (x−y)= 2 f (x) f (y), for all x,y ∈ R;
(b) there exists x0 with f (x0) = −1.
Prove that f is periodic.
3. If a function f : R→R is not injective and there exists a function g : R×R→R
such