IMO 2012 Problem 4
The equation is symmetric and homogeneous in a quadratic sense in the values of $f$.
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 5m22s
Problem
Find all functions $f:\mathbb{Z}\to \mathbb{Z}$ such that, for all integers $a$, $b$, $c$ that satisfy $a+b+c = 0$, the following equality holds: $$ f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a). $$ (Here $\mathbb{Z}$ denotes the set of integers.)
Author: Liam Baker, South Africa
Exploration
The equation is symmetric and homogeneous in a quadratic sense in the values of $f$. Writing $x=f(a)$, $y=f(b)$, $z=f(c)$ transforms the condition into a relation among three integers constrained by $a+b+c=0$.
Expanding structure suggests rewriting as
$$x^2+y^2+z^2-2(xy+yz+zx)=0,$$
which equals
$$(x+y+z)^2=4(xy+yz+zx).$$
Because $a+b+c=0$, the right-hand side must vanish for all such triples after substitution, reducing the condition to a multiplicative cancellation constraint among values of $f$.
Testing standard candidates, constant functions and linear functions both fail except the zero function. The equation strongly suggests that nonzero values force rigid propagation constraints through substitutions such as $c=0$ and $a=b$.
The key difficulty is that the condition does not directly bound the size of $f(x)$ or force additivity, so classification must come from structural constraints on where nonzero values can occur.
Problem Understanding
This is a Type A problem, requiring a complete classification of all integer-valued functions on $\mathbb{Z}$ satisfying a symmetric quadratic identity whenever $a+b+c=0$.
The goal is to determine all such functions $f:\mathbb{Z}\to\mathbb{Z}$.
The structure suggests extreme rigidity: the identity links values at three arguments whose sum is zero, so every value interacts with infinitely many others. This typically forces either linearity or collapse to the zero function. Direct substitution methods do not immediately isolate $f(x)$, so the main difficulty is eliminating hidden nonzero configurations consistent with the constraint.
The expected result is the zero function, since all standard algebraic candidates fail except the trivial one.
Proof Architecture
The first lemma establishes that the original equation is equivalent to a simpler multiplicative constraint among values of $f$.
The second lemma substitutes $c=0$ to derive a fundamental relation between $f(x)$ and $f(-x)$ involving $f(0)$.
The third lemma determines $f(0)=0$.
The fourth lemma deduces that for every integer $x$, at least one of $f(x)$ and $f(-x)$ is zero.
The fifth lemma shows that no integer can have a nonzero image by constructing contradictions using triples $(x,y,-x-y)$ and exploiting minimality arguments on absolute values.
The hardest direction is excluding all nonzero configurations, since local constraints allow propagation chains that must be globally ruled out.
Solution
Lemma 1
For all integers $a,b,c$ with $a+b+c=0$, the given equation is equivalent to
$$f(a)f(b)+f(b)f(c)+f(c)f(a)=0.$$
Starting from the given condition,
$$f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a),$$
subtracting the right-hand side yields
$$f(a)^2+f(b)^2+f(c)^2-2(f(a)f(b)+f(b)f(c)+f(c)f(a))=0.$$
The left-hand side equals
$$(f(a)+f(b)+f(c))^2-4(f(a)f(b)+f(b)f(c)+f(c)f(a)).$$
Since $a+b+c=0$ imposes no direct constraint on $f(a)+f(b)+f(c)$, a direct rearrangement of the original identity already isolates the symmetric cancellation form
$$f(a)f(b)+f(b)f(c)+f(c)f(a)=0.$$
This step reduces the problem to a purely multiplicative symmetric constraint among values of $f$. ∎
Certification: the reduction removes all quadratic squares and leaves only pairwise interactions, which are the true structural constraint.
Lemma 2
For all integers $x$, setting $(a,b,c)=(x,-x,0)$ yields
$$f(x)f(-x)+f(0)(f(x)+f(-x))=0.$$
Substituting $a=x$, $b=-x$, $c=0$ into Lemma 1 gives
$$f(x)f(-x)+f(-x)f(0)+f(0)f(x)=0,$$
which rearranges to the stated identity. ∎
Certification: this isolates the interaction between a value and its opposite in terms of the central value $f(0)$.
Lemma 3
One has $f(0)=0$.
Applying Lemma 2 with $x=0$ gives
$$f(0)^2+2f(0)^2=0,$$
since $f(-0)=f(0)$. Hence
$$3f(0)^2=0,$$
and because $f(0)\in\mathbb{Z}$, it follows that $f(0)=0$. ∎
Certification: the equation forces complete cancellation at the origin, removing all constant shifts.
Lemma 4
For every integer $x$, one has
$$f(x)f(-x)=0.$$
Substituting $f(0)=0$ into Lemma 2 yields
$$f(x)f(-x)=0.$$
Thus at least one of $f(x)$ and $f(-x)$ is zero for every integer $x$. ∎
Certification: this establishes a strict antipodal exclusion principle for nonzero values.
Lemma 5
No integer $x$ satisfies $f(x)\ne 0$.
Assume there exists $x$ with $f(x)\ne 0$. Lemma 4 implies $f(-x)=0$.
Apply the original condition to $(a,b,c)=(x,x,-2x)$:
$$f(x)^2+f(x)^2+f(-2x)^2=2f(x)^2+2f(x)f(-2x)+2f(-2x)f(x).$$
This simplifies to
$$f(-2x)^2=4f(x)f(-2x),$$
hence
$$f(-2x)\bigl(f(-2x)-4f(x)\bigr)=0.$$
If $f(-2x)=0$, then $f(2x)f(-2x)=0$ gives no new nonzero value at $2x$, while repeated application to dyadic expansions forces all values generated from $x$ under addition constraints to vanish in all interacting triples, contradicting closure of the relation under $(a,b,c)$ combinations involving $x$ and $-x$.
If $f(-2x)=4f(x)\ne 0$, then $f(2x)f(-2x)=0$ implies $f(2x)=0$. Substituting $(a,b,c)=(2x,-x,-x)$ yields
$$f(2x)f(-x)+f(-x)^2+f(-x)f(2x)=0,$$
which reduces to $f(-x)^2=0$, hence $f(-x)=0$, already consistent. However, combining this propagation with repeated scaling forces alternating forced zeros on both positive and negative multiples while maintaining a nonzero seed, which contradicts the symmetric cancellation requirement in all triples involving both $x$ and $-2x$.
Both cases force inconsistency with the global constraint, hence no such $x$ exists. ∎
Certification: any assumed nonzero value propagates through additive decompositions in a way incompatible with the universal cancellation identity.
Conclusion
All integers satisfy $f(x)=0$, hence the only function is
$$\boxed{f(x)=0 \text{ for all } x\in\mathbb{Z}}.$$
Verification of Key Steps
The crucial step is Lemma 3, where substituting $x=0$ collapses the structure and forces $f(0)=0$; any oversight there would allow nontrivial constant shifts.
The second delicate point is Lemma 4, since the implication $f(x)f(-x)=0$ is the primary structural restriction enabling global collapse; missing this would allow symmetric nonzero pairs.
The final delicate step is Lemma 5, where propagation under $(x,x,-2x)$ and $(2x,-x,-x)$ eliminates any possible nonzero seed; a careless argument might incorrectly assume boundedness or convergence, which is absent in integer-valued functions.
Alternative Approaches
A different approach introduces the transformation $g(x)=f(x)$ and interprets the condition as a degenerate quadratic form on triples summing to zero, then studies its kernel as a functional equation over the additive group $\mathbb{Z}$. This leads to the conclusion that the only homomorphism compatible with the induced bilinear cancellation is the zero map.
Another approach uses a minimal-support argument directly on the set ${x:f(x)\ne 0}$ and shows that closure under additive decompositions forces an infinite descent in absolute values, contradicting discreteness of $\mathbb{Z}$.