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Sweden Contests
Swedish Mathematical Competition
1978 Swedish Mathematical Competition
6
6
Part of
1978 Swedish Mathematical Competition
Problems
(1)
p'(x) = nq(x) if p(x)^2 = (x^2 - 1)q(x)^2 + 1
Source: 1978 Swedish Mathematical Competition p6
3/26/2021
p
(
x
)
p(x)
p
(
x
)
is a polynomial of degree
n
n
n
with leading coefficient
c
c
c
, and
q
(
x
)
q(x)
q
(
x
)
is a polynomial of degree
m
m
m
with leading coefficient
c
c
c
, such that
p
(
x
)
2
=
(
x
2
−
1
)
q
(
x
)
2
+
1
p(x)^2 = \left(x^2 - 1\right)q(x)^2 + 1
p
(
x
)
2
=
(
x
2
−
1
)
q
(
x
)
2
+
1
Show that
p
′
(
x
)
=
n
q
(
x
)
p'(x) = nq(x)
p
′
(
x
)
=
n
q
(
x
)
.
analysis
algebra
polynomial