A Functional Equation from the 2012 IMO

Here's Problem 4 from the 2012 IMO:

Find all functions \( f:\mathbb Z\to\mathbb Z \) such that, for all integers \( a \), \( b \), \( c \) with \( a+b+c=0 \) the following equality holds:


Here \( \mathbb Z\) is the set of all integers.


The problem is reasonably standard. Setting first \(a = b = c = 0\), we get

\( 3f^{2}(0)=6f^{2}(0),\)

implying \(f(0)=0\).

Next, let say \(c=0\), so that \(b=-a\): \( f^{2}(a)+f^{2}(-a)=2f(a)f(-a)\). In other words,

\((f(a) - f(-a))^{2}=0\),

so that \(f(a)=f(-a)\), making \(f\) an even function. Using that and setting \(a=b=1\) and \(c=-2\) gives

\(2f^{2}(1) + f^{2}(2)=4f(2)f(1) + 2f^{2}(1)\),

which is equivalent to


Thus one of the two: either \(f(2)=0\) or \(f(2)=4f(1)\). We'll have to consider the two cases separately.

Note that we can safely assume that \(f(1)\ne 0\), for $f(0)=0$ would imply that \(f\) is identically \(0\), which is an obvious solution. To see that, take successively \(a=b=1\), \(c=2\) which gives \(f(2)=0\), and then \(a=1\), \(b=2\), \(c=3\), to obtain \(f(3)=0\), and so on.

Case 1: \(f(2)=0\)

Clearly also \(f(4)=0\) and \(f(6)=0\), and so on. More generally, \(f(2n)=0\), for all \(n\in \mathbb Z\). This gives a good reason to suspect that this is an indication that \(f\) is periodic. Let's check:


But, since \(f(2)=0\), this reduces to

\(f^{2}(a+2)+f^{2}(a)=2f(a+2)f(a)\) or \((f(a+2)-f(a))^{2}=0\),

Telling us that \(f\) is periodic, with period \(2\). This means that \(f(2n)=0\) and \(f(2n+1)=k=\mbox{const}\), for all \(n\in \mathbb Z\).

Case 2: \(f(2)=4f(1)\)

Let's see what comes next:


This simplifies to


This quadratic equation yields two solutions: \(f(3)=9f(1)\) or \(f(3)=f(1)\). Again, we are forced to consider two cases separately:

Case 2.1: \(f(3)=9f(1)\)

We'll prove by induction that, for all \(n\in \mathbb Z\), \(f(n)=n^{2}f(1)\). We already saw that this is true for \(n=1,2,3\). Assuming this is true for \(n=m\),


By the inductive assumption, this reduces to

\( \begin{align} f^{2}(m+1)+(m^{4}+1)f^{2}(1) &= 2f(m+1)m^{2}f(1)+2m^{2}f^{2}(1)+2f(1)f(m+1)\\ &=2m^{2}f^{2}(1)+2f(m+1)f(1)(m^{2}+1). \end{align} \)

And, in turn, to

\(f^{2}(m+1)-2(m^{2}+1)f(m+1)f(1)+(m^{2}-1)^{2}f^{2}(1) = 0.\)

Strangely enough, solving this quadratic equation leads to \(f(m+1)=(m\pm 1)^{2}f(1)\).

Case 2.1.1: \(f(m+1)=(m+1)^{2}f(1)\)

The "\(+\)" sign confirms the expected result. It is easily checked that \(f(n)=n^{2}f(1)\) serves a solution to the given equation.

Case 2.1.2: \(f(m+1)=(m-1)^{2}f(1)\)

This case turns up rather surprisingly. We started with the sequence \(f(1)\), \(f(2)=2^{2}f(1)\), \(f(3)=3^{2}f(1)\), but along the way there emerged a possibility that for some \(m\) \(f(m+1)=(m-1)^{2}f(1)\). Assuming this to be the case, let \(k\gt 2\) be the least such number. Then in particular,

\( \begin{align} f(k-1)&=(k-1)^{2}f(1) \\ f(k+1)&=(k-1)^{2}f(1) \end{align} \)

Substituting this into


and dropping the common factor \(f^{2}(1)\), gives

\( (k-1)^{4} + (k-1)^{4} + 4^{2} = 2(k-1)^{2}(k-1)^{2} + 2\cdot 4\cdot (k-1)^{2} + 2\cdot 4\cdot (k-1)^{2}, \)

which simplifies to

\( 1 = (k-1)^{2}. \)

This equation has two solutions: \(k=0\) and \(k=2\), neither of which fits the Case 2.1 requirement \(k\gt 2\). Thus this pass does not lead to a solution of the problem.

Case 2.2: \(f(3)=f(1)\)

Let's find \(f(4)\) in this case:




which is the same as


so that either \(f(4)=0\) or \(f(4)=4f(1)\). Two cases again:

Case 2.2.1: \(f(4)=0\)

Repeating Case 1, we see that \(f\) is periodic with period \(4\). The values \(f(1),f(2),f(3),f(4)\) are \(f(1), 4f(1), f(1), 0\), respectively, and these are repeated.

Case 2.2.2: \(f(4)=4f(1)\)

Since also \(f(2)=4f(1)\), \(f(4)=f(2)\). But then,


simplifies to


implying \(f(2)=0\), which is Case 1.

Summing up, there are three kinds of solutions:

  1. \(f(2n)=0\), \(f(2n+1)=C,\)
  2. \(f(n)=n^{2}C,\)
  3. \(f(4n+1)=f(4n+3)=C\), \(f(4n+2)=4C\), \(f(4n)=0\),

where \(C\) is an arbitrary (integer) constant, \(n\in \mathbb Z\).

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