Subfield: Algebra


Question: Find all the functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that

$$
f(a)^{2}+f(b)^{2}+f(c)^{2}=2 f(a) f(b)+2 f(b) f(c)+2 f(c) f(a)
$$

for all integers $a, b, c$ satisfying $a+b+c=0$.

Solution:
The substitution $a=b=c=0$ gives $3 f(0)^{2}=6 f(0)^{2}$, hence

$$
f(0)=0 \text {. }\tag{1}
$$

The substitution $b=-a$ and $c=0$ gives $\left((f(a)-f(-a))^{2}=0\right.$. Hence $f$ is an even function:

$$
f(a)=f(-a) \quad \text { for all } a \in \mathbb{Z}\tag{2}
$$

Now set $b=a$ and $c=-2 a$ to obtain $2 f(a)^{2}+f(2 a)^{2}=2 f(a)^{2}+4 f(a) f(2 a)$. Hence

$$
f(2 a)=0 \text { or } f(2 a)=4 f(a) \text { for all } a \in \mathbb{Z}\tag{3}
$$

If $f(r)=0$ for some $r \geq 1$ then the substitution $b=r$ and $c=-a-r$ gives $(f(a+r)-f(a))^{2}=0$. So $f$ is periodic with period $r$, i. e.

$$
f(a+r)=f(a) \text { for all } a \in \mathbb{Z}
$$

In particular, if $f(1)=0$ then $f$ is constant, thus $f(a)=0$ for all $a \in \mathbb{Z}$. This function clearly satisfies the functional equation. For the rest of the analysis, we assume $f(1)=k \neq 0$.

By (3) we have $f(2)=0$ or $f(2)=4 k$. If $f(2)=0$ then $f$ is periodic of period 2 , thus $f($ even $)=0$ and $f($ odd $)=k$. This function is a solution for every $k$. We postpone the verification; for the sequel assume $f(2)=4 k \neq 0$.

By (3) again, we have $f(4)=0$ or $f(4)=16 k$. In the first case $f$ is periodic of period 4 , and $f(3)=f(-1)=f(1)=k$, so we have $f(4 n)=0, f(4 n+1)=f(4 n+3)=k$, and $f(4 n+2)=4 k$ for all $n \in \mathbb{Z}$. This function is a solution too, which we justify later. For the rest of the analysis, we assume $f(4)=16 k \neq 0$.

We show now that $f(3)=9 k$. In order to do so, we need two substitutions:

$$
\begin{gathered}
a=1, b=2, c=-3 \Longrightarrow f(3)^{2}-10 k f(3)+9 k^{2}=0 \Longrightarrow f(3) \in\{k, 9 k\} \\
a=1, b=3, c=-4 \Longrightarrow f(3)^{2}-34 k f(3)+225 k^{2}=0 \Longrightarrow f(3) \in\{9 k, 25 k\}
\end{gathered}
$$

Therefore $f(3)=9 k$, as claimed. Now we prove inductively that the only remaining function is $f(x)=k x^{2}, x \in \mathbb{Z}$. We proved this for $x=0,1,2,3,4$. Assume that $n \geq 4$ and that $f(x)=k x^{2}$ holds for all integers $x \in[0, n]$. Then the substitutions $a=n, b=1, c=-n-1$ and $a=n-1$, $b=2, c=-n-1$ lead respectively to

$$
f(n+1) \in\left\{k(n+1)^{2}, k(n-1)^{2}\right\} \quad \text { and } \quad f(n+1) \in\left\{k(n+1)^{2}, k(n-3)^{2}\right\}
$$

Since $k(n-1)^{2} \neq k(n-3)^{2}$ for $n \neq 2$, the only possibility is $f(n+1)=k(n+1)^{2}$. This completes the induction, so $f(x)=k x^{2}$ for all $x \geq 0$. The same expression is valid for negative values of $x$ since $f$ is even. To verify that $f(x)=k x^{2}$ is actually a solution, we need to check the identity $a^{4}+b^{4}+(a+b)^{4}=2 a^{2} b^{2}+2 a^{2}(a+b)^{2}+2 b^{2}(a+b)^{2}$, which follows directly by expanding both sides.



Therefore the only possible solutions of the functional equation are the constant function $f_{1}(x)=0$ and the following functions:

$$
f_{2}(x)=k x^{2} \quad f_{3}(x)=\left\{\begin{array}{ccc}
0 & x \text { even } \\
k & x \text { odd }
\end{array} \quad f_{4}(x)=\left\{\begin{array}{ccc}
0 & x \equiv 0 & (\bmod 4) \\
k & x \equiv 1 & (\bmod 2) \\
4 k & x \equiv 2 & (\bmod 4)
\end{array}\right.\right.
$$

for any non-zero integer $k$. The verification that they are indeed solutions was done for the first two. For $f_{3}$ note that if $a+b+c=0$ then either $a, b, c$ are all even, in which case $f(a)=f(b)=f(c)=0$, or one of them is even and the other two are odd, so both sides of the equation equal $2 k^{2}$. For $f_{4}$ we use similar parity considerations and the symmetry of the equation, which reduces the verification to the triples $(0, k, k),(4 k, k, k),(0,0,0),(0,4 k, 4 k)$. They all satisfy the equation.

Final answer: $f_{1}(x)=0$ and the following functions:

$$
f_{2}(x)=k x^{2} \quad f_{3}(x)=\left\{\begin{array}{ccc}
0 & x \text { even } \\
k & x \text { odd }
\end{array} \quad f_{4}(x)=\left\{\begin{array}{ccc}
0 & x \equiv 0 & (\bmod 4) \\
k & x \equiv 1 & (\bmod 2) \\
4 k & x \equiv 2 & (\bmod 4)
\end{array}\right.\right.
$$