Difference between revisions of "1975 Canadian MO Problems/Problem 8"
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In the first case, <math>P(x)</math> is a constant. This means that <math>c = c^k \Longrightarrow c\in \{0,1\}</math> or <math>c\in \{0,1,-1\}</math> if <math>k</math> is even. | In the first case, <math>P(x)</math> is a constant. This means that <math>c = c^k \Longrightarrow c\in \{0,1\}</math> or <math>c\in \{0,1,-1\}</math> if <math>k</math> is even. | ||
− | In the second case, <math>P(x)</math> is nonconstant with coefficients of <math>a_1,a_2,\hdots a_{k+1}</math>. If we divide by <math>P(x)</math> on both sides, then we have that <math>a_1 + \frac{a_2}{P(x)} + \frac{ | + | In the second case, <math>P(x)</math> is nonconstant with coefficients of <math>a_1,a_2,\hdots a_{k+1}</math>. If we divide by <math>P(x)</math> on both sides, then we have that <math>a_1 + \frac{a_2}{P(x)} + \frac{a_3}{P(x)^2} \hdots \frac{a_{k+1}}{P(x)^{k+1}} = 1</math>. This can only be achieved if <math>a_2,a_3, \hdots a_{k+1} = 0</math>. This is because if we factor out a <math>\frac{1}{P(x)}</math>, then clearly these terms are not constant. Thus, <math>a_1 = 1</math> and our second solution is <math>P(x) = x^k</math>. |
~bigbrain123 | ~bigbrain123 |
Revision as of 18:36, 9 June 2022
Problem 8
Let be a positive integer. Find all polynomials
where the
are real, which satisfy the equation
.
1975 Canadian MO (Problems) | ||
Preceded by Problem 7 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by "Last Question" |
Solution 1
Let be the degree of polynomial
. We begin by noting that
. This is because the degree of the LHS is
and the RHS is
. Now we split
into two cases.
In the first case, is a constant. This means that
or
if
is even.
In the second case, is nonconstant with coefficients of
. If we divide by
on both sides, then we have that
. This can only be achieved if
. This is because if we factor out a
, then clearly these terms are not constant. Thus,
and our second solution is
.
~bigbrain123