Mathematics Olympiad Coachs Seminar, Zhuhai, China 1
03/22/2004
Algebra
1.
There exists a polynomial P of degree 5 with the following property: if z is a complex number such
that z5 + 2004z = 1, then P(z2) = 0.
Find all such polynomials P.
2.
Let N...
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Mathematics Olympiad Coachs Seminar, Zhuhai, China 1
03/22/2004
Algebra
1.
There exists a polynomial P of degree 5 with the following property: if z is a complex number such
that z5 + 2004z = 1, then P(z2) = 0.
Find all such polynomials P.
2.
Let N denote the set of positive integers.
Find all functions f : N → N such that
f(m + n)f(m − n) = f(m2
)
for all m, n ∈ N.
Solution: Function f(n) = 1, for all n ∈ N, is the only function satisfying the conditions of
the problem.
Note that
f(1)f(2n − 1) = f(n2
) and f(3)f(2n − 1) = f((n + 1)2
)
for n ≥ 3.
Thus
f(3)
f(1)
=
f((n + 1)2)
f(n2)
.
Setting f(3)
f(1) = k yields f(n2) = kn−3f(9) for n ≥ 3.
Similarly, for all h ≥ 1,
f(h + 2)
f(h)
=
f((m + 1)2)
f(m2)
for sufficiently large m and is thus also k.
Hence f(2h) = kh−1f(2) and f(2h + 1) = khf(1).
But
f(25)
f(9)
=
f(25)
f(23)
· · · · ·
f(11)
f(9)
= k8
and
f(25)
f(9)
=
f(25)
f(16)
·
f(16)
f(9)
= k2
,
so k = 1 and f(16) = f(9).
This implies that f(2h + 1) = f(1) = f(2) = f(2j) for all j
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