Difference between revisions of "1975 Canadian MO Problems/Problem 8"
Bigbrain123 (talk | contribs) (→Problem 8: Wrote a solution.) |
Bigbrain123 (talk | contribs) (→Solution 1) |
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== Solution 1== | == Solution 1== | ||
− | Let <math>f(n)</math> be the degree of polynomial <math>n</math>. We begin by noting that <math>f(P(x)) = k</math>. This is because the degree of the LHS is <math>f(P(x))^f(P(x))</math> and the RHS is <math>f(P(x))^k</math>. Now we split <math>P(x)</math> into two cases. | + | Let <math>f(n)</math> be the degree of polynomial <math>n</math>. We begin by noting that <math>f(P(x)) = k</math>. This is because the degree of the LHS is <math>f(P(x))^{f(P(x))}</math> and the RHS is <math>f(P(x))^k</math>. Now we split <math>P(x)</math> into two cases. |
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. |
Revision as of 22:09, 29 July 2021
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