Advanced book on Mathematics Olympiad

Algebra 457

is true ifBCis replaced byAB. It follows thatis preserved both by the translations in
the groupGBCand in the analogous groupGAB. These generate a group that is dense in
the group of all translations of the plane. We conclude thatis a dense set in the plane,
as desired.
(communicated by K. Shankar)

290.The symmetry groups are, respectively,C 2 v,D 2 h, andD 2 d.

291.Ifxis an idempotent, then 1−xis an idempotent as well. Indeed,

( 1 −x)^2 = 1 − 2 x+x^2 = 1 − 2 x+x= 1 −x.

Thus there is an involution onM,x → 1 −x. This involution has no fixed points,
sincex= 1 −ximpliesx^2 =x−x^2 orx=x−x=0. But then 0= 1 − 0 =1,
impossible. Having no fixed points, the involution pairs the elements ofM, showing that
the cardinality ofMis even.
(Gazeta Matematica ̆(Mathematics Gazette, Bucharest), proposed by V. Zidaru)

292.We havey=y^6 =(−y)^6 =−y, hence 2y=0 for anyy∈R. Now letxbe an
arbitrary element inR. Using the binomial formula, we obtain

x+ 1 =(x+ 1 )^6 =x^6 + 6 x^5 + 15 x^4 + 20 x^3 + 15 x^2 + 6 x+ 1

=x^4 +x^2 +x+ 1 ,

where we canceled the terms that had even coefficients. Hencex^4 +x^2 =0, orx^4 =
−x^2 =x^2. We then have

x=x^6 =x^2 x^4 =x^2 x^2 =x^4 =x^2 ,

and sox^2 =x, as desired. From the equality(x+y)^2 =x+ywe deducexy+yx=0,
soxy=−yx=yxfor anyx, y. This shows that the ring is commutative, as desired.

293.Substitutingxbyx+1 in the relation from the statement, we find that

((x+ 1 )y)^2 −(x+ 1 )^2 y^2 =(xy)^2 +xy^2 +yxy+y^2 −x^2 y^2 − 2 xy^2 −y^2

=yxy−xy^2 = 0.

Hencexy^2 =yxyfor allx, y∈R. Substituting in this relationybyy+1, we find that

xy^2 + 2 xy+x=yxy+yx+xy+x.

Using the fact thatxy^2 =yxy, we obtainxy=yx, as desired.

294.This problem generalizes the first example from the introduction. The idea of the
solution is similar. Now letvbe the inverse of 1−(xy)n. Thenv( 1 −(xy)n) =