Advanced book on Mathematics Olympiad

Algebra 455

M^2 =(G 1 +G 2 +···+Gk)^2 =

∑k

i= 1

Gi

⎛

⎝

∑k

j= 1

Gj

⎞

⎠=

∑k

i= 1

Gi

(

∑

G∈

G−i^1 G

)

=

∑

G∈

∑k

i= 1

Gi(G−i^1 G)=k

∑

G∈

G=kM.

Taking determinants, we find that(detM)^2 =kndetM. Hence either detM =0or

detMis equal to the order ofraised to thenth power.

Remark.In fact, much more is true. The determinant of the sum of the elements of a
finite multiplicative group of matrices is nonzero only when the group consists of one
element, the identity, in which case it is equal to 1. This is the corollary of a basic fact in
representation theory.
A representation of a group is a homomorphism of the group into a group of matrices.
In our situation the group is already represented as a group of matrices. A representation is
called irreducible if there does not exist a basis in which it can be decomposed into blocks.
Any representation of a finite group is the block sum of irreducible representations. The
simplest representation, called the trivial representation, sends all elements of the group
to the identity element. A result in representation theory states that for any nontrivial
irreducible representation of a finite group, the sum of the matrices of the representation
is zero. In an appropriately chosen basis, our group can be written as the block sum of
irreducible representations. If the group is nontrivial, then at least one representation is
nontrivial. In summing the elements of the group, the diagonal block corresponding to
this irreducible representation is the zero matrix. Taking the determinant, we obtain zero.

285.The condition from the statement implies that for all integersmandn,

f(m

√

2 +n

√

3 )=f( 0 ).

Because the ratio

√

2 /

√

3 is irrational, the additive group generated by

√

2 and

√

3isnot

cyclic. It means that this group is dense inR.Sofis constant on a dense subset ofR.

Being continuous, it must be constant on the real axis.

286.The conclusion follows from the fact that the additive group

S={n+ 2 πm;m, nintegers}

is dense in the real numbers. Indeed, by the result we just proved, we only need to check

thatSis not cyclic. This is so becausenand 2mπcannot both be integer multiples of

the same number (they are incommensurable).

287.That 2kstarts witha7isequivalent to the existence of an integermsuch^2

k

10 m∈[^7 ,^8 ).

Let us show that the set{^2

k

10 m | k, mintegers}is dense in the positive real numbers.