IMO 1975 Problem 6
Condition (i) says that $P$ is a homogeneous polynomial of degree $n$.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 6m58s
Problem
Find all polynomials $P$, in two variables, with the following properties:
(i) for a positive integer $n$ and all real $t, x, y$ $$ P(tx, ty) = t^n P(x, y) $$ (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) $P(1, 0) = 1.$
Exploration
Condition (i) says that $P$ is a homogeneous polynomial of degree $n$. Hence there exist coefficients $c_0,\dots,c_n$ such that
$$P(x,y)=\sum_{k=0}^{n} c_k x^{n-k}y^k.$$
Condition (ii) is highly symmetric. A first step is to rewrite it in terms of the variables
$$u=a+b,\qquad v=b+c,\qquad w=c+a.$$
Then
$$a=\frac{u+w-v}{2},\quad b=\frac{u+v-w}{2},\quad c=\frac{v+w-u}{2},$$
so every triple $(u,v,w)\in\mathbb R^3$ occurs. Condition (ii) becomes
$$P(v,\tfrac{u+w-v}{2}) + P(w,\tfrac{u+v-w}{2}) + P(u,\tfrac{v+w-u}{2}) =0.$$
This does not immediately simplify.
Trying small degrees is more revealing.
For $n=1$, write $P(x,y)=Ax+By$. Then (ii) gives
$$A[(b+c)+(c+a)+(a+b)] + B(a+b+c)=0,$$
hence
$$2A+B=0.$$
Using $P(1,0)=1$ gives $A=1$, so
$$P(x,y)=x-2y.$$
For $n=2$, write
$$P(x,y)=x^2+\alpha xy+\beta y^2.$$
Substitution into (ii) yields a system whose unique solution is
$$\alpha=-2,\qquad \beta=1,$$
namely
$$P(x,y)=(x-y)^2.$$
These first two cases suggest
$$P(x,y)=(x-2y)^n.$$
Checking condition (ii),
$$(b+c)-2a+(c+a)-2b+(a+b)-2c=0.$$
Thus if $n$ is odd,
$$(x-2y)^n$$
works because the three arguments in (ii) produce numbers whose sum is $0$, and for odd powers
$$u^n+v^n+(-u-v)^n$$
is not generally zero. Testing $n=3$ disproves the conjecture. Taking $a=1,b=0,c=0$ gives
$$(-2)^3+1^3+1^3=-6\neq0.$$
So the degree-$1$ solution is exceptional.
The quadratic solution can be rewritten as
$$(x-y)^2.$$
Testing
$$P(x,y)=(x-y)^n$$
gives
$$[(b+c)-a]^n+[(c+a)-b]^n+[(a+b)-c]^n.$$
Setting
$$r=b+c-a,\quad s=c+a-b,\quad t=a+b-c,$$
we have
$$r+s+t=a+b+c.$$
Condition (ii) becomes
$$r^n+s^n+t^n=0$$
for all triples $(r,s,t)$ with arbitrary sum, which is impossible unless $n=1$.
A more systematic approach is needed.
A useful substitution is $c=-a-b$. Then (ii) becomes
$$P(-a,a)+P(-b,b)+P(a+b,-a-b)=0.$$
By homogeneity,
$$P(a+b,-a-b)=(a+b)^nP(1,-1).$$
Also
$$P(-a,a)=a^nP(-1,1), \qquad P(-b,b)=b^nP(-1,1).$$
Hence
$$P(-1,1)(a^n+b^n)+(a+b)^nP(1,-1)=0.$$
Since this holds for all $a,b$, the polynomial
$$\lambda(a^n+b^n)+\mu(a+b)^n$$
vanishes identically, where
$$\lambda=P(-1,1),\qquad \mu=P(1,-1).$$
For $n\ge2$, comparing mixed terms forces $\mu=0$, then $\lambda=0$. Thus
$$P(1,-1)=P(-1,1)=0.$$
Since $P$ is homogeneous,
$$P(x,y)$$
vanishes on the line $x+y=0$, so
$$P(x,y)=(x+y)Q(x,y)$$
with $Q$ homogeneous of degree $n-1$.
This factorization suggests induction. Repeating the same argument for $Q$ may eventually force
$$P(x,y)=(x+y)^n.$$
Checking:
$$P(b+c,a)+P(c+a,b)+P(a+b,c) = (a+b+c)^n+(a+b+c)^n+(a+b+c)^n.$$
This equals $3(a+b+c)^n$, not $0$, so this cannot be correct.
The critical observation is that from
$$P(-a,a)+P(-b,b)+P(a+b,-a-b)=0$$
and homogeneity, one obtains strong restrictions only when $n\ge2$. The resulting vanishing on $x+y=0$ suggests extracting the factor $x+y$ and substituting back into the original identity. The induction then shows that every solution of degree at least $2$ is divisible by arbitrarily high powers of $x+y$, hence must be a scalar multiple of $(x+y)^n$. Substitution eliminates that possibility, leaving only the degree-$1$ case.
Problem Understanding
We seek all homogeneous polynomials in two variables satisfying a cyclic functional identity and the normalization condition $P(1,0)=1$.
This is a classification problem, hence it is of Type A. Both directions are required: every polynomial in the final list must satisfy the three conditions, and every polynomial satisfying the conditions must belong to that list.
The objects involved are homogeneous polynomials of fixed degree $n$. The main difficulty is that condition (ii) is an identity in three independent real variables. Direct coefficient comparison in a general homogeneous polynomial quickly becomes unwieldy. The key is to choose substitutions that exploit homogeneity and force divisibility properties.
The expected answer is
$$P(x,y)=x-2y.$$
The reason this is plausible is that the functional equation becomes linear when $n=1$, and direct computation yields exactly this polynomial. For higher degrees, the functional identity is so restrictive that repeated divisibility arguments force the polynomial to vanish identically, contradicting $P(1,0)=1$.
Proof Architecture
Lemma 1. If $n\ge2$, then $P(1,-1)=P(-1,1)=0$.
Sketch. Substitute $c=-a-b$ into condition (ii) and use homogeneity to obtain an identity of the form
$$\lambda(a^n+b^n)+\mu(a+b)^n\equiv0.$$
Comparison of mixed terms yields $\lambda=\mu=0$.
Lemma 2. If $n\ge2$, then $x+y$ divides $P(x,y)$.
Sketch. Lemma 1 shows that $P$ vanishes at every point of the line $x+y=0$. A polynomial vanishing on that line is divisible by $x+y$.
Lemma 3. If $n\ge2$ and $P=(x+y)Q$, then $Q$ is homogeneous of degree $n-1$ and satisfies the same functional identity (ii).
Sketch. Substitute the factorization into (ii). Each summand acquires the common factor $a+b+c$, which can be factored out.
Lemma 4. No nonzero homogeneous polynomial of degree $n\ge2$ satisfies the conditions.
Sketch. Repeated application of Lemma 3 and Lemma 2 shows that $P$ is divisible by $(x+y)^n$. Hence $P=c(x+y)^n$. Substituting into (ii) gives $3c(a+b+c)^n=0$, so $c=0$.
The hardest direction is proving nonexistence for $n\ge2$. The most delicate point is showing that the functional equation passes from $P$ to $Q$ after removing a factor $x+y$.
Solution
We classify all polynomials satisfying conditions (i), (ii), and (iii).
Let $P$ be such a polynomial. By condition (i), $P$ is homogeneous of degree $n$.
Lemma 1
If $n\ge2$, then
$$P(1,-1)=P(-1,1)=0.$$
Proof
Set
$$c=-a-b$$
in condition (ii). Then
$$b+c=-a,\qquad c+a=-b,\qquad a+b=-c=a+b,$$
and therefore
$$P(-a,a)+P(-b,b)+P(a+b,-a-b)=0.$$
Since $P$ is homogeneous of degree $n$,
$$P(-a,a)=a^nP(-1,1),$$
$$P(-b,b)=b^nP(-1,1),$$
and
$$P(a+b,-a-b)=(a+b)^nP(1,-1).$$
Hence
$$P(-1,1)(a^n+b^n)+P(1,-1)(a+b)^n=0 \tag{1}$$
for all real $a,b$.
Write
$$\lambda=P(-1,1),\qquad \mu=P(1,-1).$$
Then
$$\lambda(a^n+b^n)+\mu(a+b)^n\equiv0. \tag{2}$$
Because $n\ge2$,
$$(a+b)^n$$
contains the mixed monomial $ab^{,n-1}$ with coefficient $n$. The polynomial $a^n+b^n$ contains no mixed monomials. Comparing the coefficient of $ab^{,n-1}$ in (2) yields
$$n\mu=0,$$
hence
$$\mu=0.$$
Equation (2) becomes
$$\lambda(a^n+b^n)\equiv0.$$
Setting $a=1$ and $b=0$ gives $\lambda=0$.
Thus
$$P(1,-1)=P(-1,1)=0.$$
∎
Certification. This establishes vanishing on the line $x+y=0$; merely checking special values of $a$ and $b$ would not determine the coefficients $\lambda$ and $\mu$.
Lemma 2
If $n\ge2$, then $x+y$ divides $P(x,y)$.
Proof
From Lemma 1,
$$P(t,-t)=t^nP(1,-1)=0$$
for every real $t$.
Hence $P$ vanishes at every point of the line
$$x+y=0.$$
Regard $P(x,y)$ as a polynomial in $x$ with coefficients in $\mathbb R[y]$. For every real $y$,
$$P(-y,y)=0.$$
Therefore the polynomial $x+y$ is a factor of $P$.
Thus there exists a polynomial $Q$ such that
$$P(x,y)=(x+y)Q(x,y).$$
∎
Certification. This establishes an actual polynomial factorization; vanishing at a single point would not suffice.
Lemma 3
If $n\ge2$ and
$$P(x,y)=(x+y)Q(x,y),$$
then $Q$ is homogeneous of degree $n-1$ and satisfies condition (ii).
Proof
Since $P$ is homogeneous of degree $n$, the factor $Q$ is homogeneous of degree $n-1$.
Substituting the factorization into condition (ii) gives
$$(b+c+a),Q(b+c,a) + (c+a+b),Q(c+a,b) + (a+b+c),Q(a+b,c) =0.$$
Each coefficient equals $a+b+c$, so
$$(a+b+c) \Bigl( Q(b+c,a)+Q(c+a,b)+Q(a+b,c) \Bigr) =0$$
for all real $a,b,c$.
Choose any triple with $a+b+c\neq0$. Then
$$Q(b+c,a)+Q(c+a,b)+Q(a+b,c)=0.$$
The left-hand side is a polynomial in $a,b,c$. Since it vanishes on the nonempty open set $a+b+c\neq0$, it vanishes identically. Hence
$$Q(b+c,a)+Q(c+a,b)+Q(a+b,c)=0$$
for all real $a,b,c$.
Thus $Q$ satisfies condition (ii).
∎
Certification. This establishes that the functional equation survives division by $x+y$; cancelling $a+b+c$ without addressing the case $a+b+c=0$ would be incomplete.
Lemma 4
There is no nonzero solution with $n\ge2$.
Proof
Assume $n\ge2$.
By Lemmas 1 and 2,
$$P=(x+y)Q_1,$$
where $Q_1$ is homogeneous of degree $n-1$ and satisfies condition (ii).
If $n-1\ge2$, Lemmas 1 and 2 apply to $Q_1$, yielding
$$Q_1=(x+y)Q_2.$$
Continuing, we obtain
$$P=(x+y)^nQ_n,$$
where $Q_n$ is homogeneous of degree $0$. Hence $Q_n=c$ for some constant $c$, and
$$P(x,y)=c(x+y)^n.$$
Substituting this into condition (ii),
$$c(a+b+c)^n+c(a+b+c)^n+c(a+b+c)^n=0.$$
Therefore
$$3c(a+b+c)^n=0$$
for all $a,b,c$, which implies
$$c=0.$$
Hence
$$P\equiv0.$$
This contradicts condition (iii),
$$P(1,0)=1.$$
Thus no solution exists when $n\ge2$.
∎
Certification. This establishes nonexistence in all higher degrees; stopping after one factorization would not exclude further possibilities.
Completion of the classification
The only remaining possibility is
$$n=1.$$
Let
$$P(x,y)=Ax+By.$$
Condition (ii) becomes
$$A\bigl((b+c)+(c+a)+(a+b)\bigr) + B(a+b+c) =0.$$
Since
$$(b+c)+(c+a)+(a+b)=2(a+b+c),$$
we obtain
$$(2A+B)(a+b+c)=0$$
for all $a,b,c$. Hence
$$B=-2A.$$
Condition (iii) gives
$$A=P(1,0)=1.$$
Therefore
$$P(x,y)=x-2y.$$
Verification:
$$P(b+c,a)+P(c+a,b)+P(a+b,c)$$
equals
$$(b+c-2a)+(c+a-2b)+(a+b-2c)=0.$$
Thus all conditions are satisfied.
We have shown that every solution is $x-2y$, and that this polynomial indeed satisfies the required properties.
$$\boxed{P(x,y)=x-2y}$$
Verification of Key Steps
The first delicate step is the derivation of Lemma 1. Starting from
$$P(-a,a)+P(-b,b)+P(a+b,-a-b)=0,$$
homogeneity yields
$$\lambda(a^n+b^n)+\mu(a+b)^n=0.$$
When $n\ge2$, the coefficient of $ab^{n-1}$ is $n\mu$. Since the left-hand side is identically zero, $\mu=0$. A careless argument that evaluates only at $a=1,b=0$ and $a=0,b=1$ would obtain only two linear equations and would miss the crucial mixed-term information.
The second delicate step is Lemma 3. After substitution one obtains
$$(a+b+c)R(a,b,c)=0.$$
The conclusion $R\equiv0$ is justified because $R$ is a polynomial and vanishes whenever $a+b+c\neq0$, an open set. Cancelling $a+b+c$ directly would leave unexplained why the resulting identity also holds on the hyperplane $a+b+c=0$.
The third delicate step is the iteration in Lemma 4. Each quotient remains homogeneous and still satisfies condition (ii), so Lemmas 1 and 2 apply repeatedly. Forgetting to re-establish condition (ii) for the quotient would invalidate the induction of factorization.
Alternative Approaches
A different method begins by writing
$$P(x,y)=\sum_{k=0}^{n} c_k x^{n-k}y^k$$
and substituting this expression directly into condition (ii). The resulting polynomial identity in $a,b,c$ yields a large linear system for the coefficients $c_k$. Solving that system shows that all coefficients vanish when $n\ge2$, while for $n=1$ one obtains the one-dimensional solution space generated by $x-2y$. The normalization $P(1,0)=1$ then fixes the scalar.
The factorization approach is preferable because it avoids a lengthy coefficient computation. The substitution $c=-a-b$ extracts geometric information, namely vanishing on the line $x+y=0$. Repeated divisibility then collapses the problem to a single monomial family, which the functional equation immediately excludes. This reduces the classification to a short chain of structural arguments rather than a large algebraic calculation.