Problem 4

Find all functions

f:\mathbb Z\to\mathbb Z

such that for all integers

a,b,c

with

a+b+c=0

f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a)

Step 2 of 5: A zero creates a period and a dichotomy

Previous step Next step

In plain words

Once one value vanishes, the equation propagates that value as a period; evaluating a special triple then controls the next even input.

f(a)=0\Longrightarrow f(a+b)=f(b),\qquad f(2a)\bigl(f(2a)-4f(a)\bigr)=0

Detailed analysis

If $f(a)=0$ , use $(a,b,-a-b)$ and evenness to obtain $(f(b)-f(a+b))^2=0$ , so a is a period. Using $(a,a,-2a)$ gives $f(2a)(f(2a)-4f(a))=0$ . In particular, with $a=1$ , either $f(2)=0$ or $f(2)=4f(1)$ .