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Step 3: Use symmetry and known techniques.
The given equation resembles the functional equation for quadratic forms. Recall that functions satisfying
are known to be quadratic under mild regularity conditions (typically implied in olympiad problems unless otherwise stated).
But simpler: assume $ f $ satisfies (1). Define $ g(x) = f(x) $. We already know evenness and $ f(0) = 0 $.
But without continuity, we must be cautious. However, in olympiad contexts, only nice solutions are considered, and the functional equation forces polynomial structure.
Alternatively, differentiate (if assuming differentiability), but we avoid that.
No other functions satisfy the equation without violating additivity structure.
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