Test for a job interview
Solution
Answer 8: There are exactly seven solutions.
If a triple (x,y,z) solves the system of equations, then the following holds:
|x| =
6 1, |y| =
6 1, |z| =
6 1, x := tan α (4)
(1)
⇐⇒ y =
2x
1−x2
x=tan α
=
2 tan α
1−tan2 α
= tan(2α) (5)
(2)
⇐⇒ z =
2y
1−y 2
=
5
2 tan(2α)
1−tan2 (2α)
= tan(4α) (6)
(3)
⇐⇒ x =
2z
1−z 2
=
6
2 tan(4α)
1−tan2 (4α)
= tan(8α) (7)
=
tan(8α)
(4) ∧ (7) =⇒
tan α
Because of the periodicity of the tangent function, there exists an integer n
satisfying α + nπ = 8α.
This implies α = nπ
, n ∈ Z.
7
By equation (4), x must also satisfy x = tan( nπ
), n ∈ Z.
7
We deduce the following possible values for x to consider, namely:
x = tan(0), x = tan( π7 ), x = tan( 2π
), x = tan( 3π
), x = tan( 4π
), x = tan( 5π
)
7
7
7
7
and x = tan( 6π
)).
7
1
According to the equations (4), (5) and (6) above, we get at most seven triples solving the system of equations:
), tan( 2nπ
), tan( 4nπ
)), n = 0, 1, 2, 3, 4, 5, 6
(tan( nπ
7
7
7
(8)
All seven triples of the form (8) satisfy the three equations (1), (2) and (3)
since for each triple, we have |x| =
6 1, |y| =
6 1, |z| =
6 1 and
2x
1−x2
=
2 tan( nπ
)
7
1−tan2 ( nπ
)
7
= tan( 2nπ
) = y
7
2y
1−y 2
=
)
2 tan( 2nπ
7
1−tan2 ( 2nπ
)
7
= tan( 4nπ
) = z
7
2z
1−z 2
=
2 tan( 4nπ
)
7
1−tan2 ( 4nπ
)
7
) = x
= tan( 8nπ
7
The latter holds since tan( 8nπ
) = tan( nπ
).
7
7
2