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 4 of 5: The nonperiodic branch is quadratic

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In plain words

If the period-two alternative is absent, the equation propagates the quadratic pattern from 1 to every positive integer.

f(n)=cn^2

Detailed analysis

Assume $f(2)=4f(1)$ and write $c=f(1)$ . If $c=0$ , this is the zero member of the quadratic family. For $c\ne0$ , induction in the original equation with $(a,b,c)= (1,n,-n-1)$ gives the two alternatives $f(n+1)=(n+1)^2c$ or $f(n+1)=(n-1)^2c$ . The latter, substituted with $(a,b,c)=(n+1,1-n,-2)$ , is impossible for n>2; the remaining initial case gives $f(3)=9c$ . Hence $f(n)=cn^2$ for all n, and direct substitution verifies this family.