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Difference between revisions of "1975 IMO Problems/Problem 6"
 
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==Problem==
 
Find all polynomials <math>P</math>, in two variables, with the following properties:
 
Find all polynomials <math>P</math>, in two variables, with the following properties:
  
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(iii) <cmath>P(1, 0) = 1.</cmath>
 
(iii) <cmath>P(1, 0) = 1.</cmath>
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==Solution 1==
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(i) If <math>n = 0</math> : Clearly no solution
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(ii) If <math>n = 1</math> : <math>P(x, y) = ax+by \implies</math> the identification yields directly <math>P(x,y) = x-2y</math>
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(iii) If <math>n > 1</math>, <math>a = 1, \ b = 1, \ c =-2 \implies P(2,-2)+P(-1, 1)+P(-1, 1) = 0</math> <math>\implies ((-2)^{n}+2) P(-1, 1) = 0 \implies P(-1, 1) = 0 \implies P(-y, y) = 0 \implies</math> <math>P(x,y)</math> is divisible by <math>(x+y)</math>
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It is then easy to see that <math>\frac{P(x, y)}{(x+y)}</math>, of degree <math>n-1</math> verifies all the equations.
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The only solutions are thus <math>P(x, y) = (x-2y)(x+y)^{n-1}</math>
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The above solution was posted and copyrighted by mathmanman. The original thread for this problem can be found here: [https://aops.com/community/p587109]
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==Solution 2==
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<math>F(a,a,a) \implies P(2a,a)=0 \implies (x-2y)</math> is a factor of <math>P(x,y)</math>.
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We may write <math>P(x,y)=(x-2y)Q(x,y)</math>
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<math>F(a,b,b) \implies 2P(a+b,b)+P(2b,a)=0 \implies 2(a+b-2b)Q(a+b,b)+2(b-a)Q(2b,a) \implies (a-b)(Q(a+b,b)-Q(2b,a))=0</math>
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Thus <math>Q(a+b,b)=Q(2b,a) \forall a \neq b</math>
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We may rewrite it as <math>Q(x,y)=Q(2y,x-y)=Q(2x-2y,3y-x=\cdots</math>
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<math>Q(x+d,y-d)-Q(x,y)</math> is a polynomial in <math>d</math> of degree <math>n-1</math> for any two fixed <math>x,y</math>,which has infinitely many zeroes,i.e,<math>0,2y-x,x-2y,6y-3x,\cdots</math>.Thus <math>Q(x+d,y-d)=Q(x,y)</math> holds for all <math>d</math>.In particular it holds for <math>d=y</math>,i.e, <math>Q(x+y,0)=Q(x,y)</math>.Now consider the polynomial <math>R(x,y)=Q(x+y,0)-Q(x,y)</math>.Suppose that its not the zero polynomial.Then its degree <math>d</math> is defined.With <math>t=\frac{x}{y}</math> it can be wriiten as <math>y^dS(t)=y^d(A(t)-B(t))</math>.But <math>S(t)</math> has infinitely many zeroes and this forces <math>A(t)=B(t)</math>,forcing <math>R(x,y)</math> to be a zero polynomial.Contradiction!.Thus <math>Q(x+y,0)</math> and <math>Q(x,y)</math> are identical polynomials.This forces <math>Q(x,y)=c(x+y)^{n-1}</math>.With <math>Q(1,0)=1</math> we get <math>c=1</math>.Thus
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<cmath>P(x,y)=(x-2y)(x+y)^{n-1}</cmath>
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The above solution was posted and copyrighted by JackXD. The original thread for this problem can be found here: [https://aops.com/community/p5745013]
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== See Also == {{IMO box|year=1974|num-b=5|after=Last Problem}}

Latest revision as of 15:21, 29 January 2021

Problem

Find all polynomials $P$, in two variables, with the following properties:

(i) for a positive integer $n$ and all real $t, x, y$ \[P(tx, ty) = t^nP(x, y)\] (that is, $P$ is homogeneous of degree $n$),

(ii) for all real $a, b, c$, \[P(b + c, a) + P(c + a, b) + P(a + b, c) = 0,\]

(iii) \[P(1, 0) = 1.\]

Solution 1

(i) If $n = 0$ : Clearly no solution (ii) If $n = 1$ : $P(x, y) = ax+by \implies$ the identification yields directly $P(x,y) = x-2y$ (iii) If $n > 1$, $a = 1, \ b = 1, \ c =-2 \implies P(2,-2)+P(-1, 1)+P(-1, 1) = 0$ $\implies ((-2)^{n}+2) P(-1, 1) = 0 \implies P(-1, 1) = 0 \implies P(-y, y) = 0 \implies$ $P(x,y)$ is divisible by $(x+y)$ It is then easy to see that $\frac{P(x, y)}{(x+y)}$, of degree $n-1$ verifies all the equations.

The only solutions are thus $P(x, y) = (x-2y)(x+y)^{n-1}$

The above solution was posted and copyrighted by mathmanman. The original thread for this problem can be found here: [1]

Solution 2

$F(a,a,a) \implies P(2a,a)=0 \implies (x-2y)$ is a factor of $P(x,y)$.

We may write $P(x,y)=(x-2y)Q(x,y)$

$F(a,b,b) \implies 2P(a+b,b)+P(2b,a)=0 \implies 2(a+b-2b)Q(a+b,b)+2(b-a)Q(2b,a) \implies (a-b)(Q(a+b,b)-Q(2b,a))=0$ Thus $Q(a+b,b)=Q(2b,a) \forall a \neq b$

We may rewrite it as $Q(x,y)=Q(2y,x-y)=Q(2x-2y,3y-x=\cdots$ $Q(x+d,y-d)-Q(x,y)$ is a polynomial in $d$ of degree $n-1$ for any two fixed $x,y$,which has infinitely many zeroes,i.e,$0,2y-x,x-2y,6y-3x,\cdots$.Thus $Q(x+d,y-d)=Q(x,y)$ holds for all $d$.In particular it holds for $d=y$,i.e, $Q(x+y,0)=Q(x,y)$.Now consider the polynomial $R(x,y)=Q(x+y,0)-Q(x,y)$.Suppose that its not the zero polynomial.Then its degree $d$ is defined.With $t=\frac{x}{y}$ it can be wriiten as $y^dS(t)=y^d(A(t)-B(t))$.But $S(t)$ has infinitely many zeroes and this forces $A(t)=B(t)$,forcing $R(x,y)$ to be a zero polynomial.Contradiction!.Thus $Q(x+y,0)$ and $Q(x,y)$ are identical polynomials.This forces $Q(x,y)=c(x+y)^{n-1}$.With $Q(1,0)=1$ we get $c=1$.Thus \[P(x,y)=(x-2y)(x+y)^{n-1}\]

The above solution was posted and copyrighted by JackXD. The original thread for this problem can be found here: [2]

See Also

1974 IMO (Problems) • Resources
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last Problem
All IMO Problems and Solutions
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