Problem
ALG-B3-M02-P019 Evenness from two means
A function \(\Phi:\mathbb R\to\mathbb R\) satisfies for all real \(u,v\): \[\Phi(u)+\Phi(v)=2\Phi\left(\frac{u+v}{2}\right)\Phi\left(\frac{u-v}{2}\right).\] Prove that \(\Phi\) is even.
Hint 1. Write the equation for \(u,-v\).
Hint 2. The right side will not change because the two factors only swap places.
For \(u,v\), \(\Phi(u)+\Phi(v)=2\Phi((u+v)/2)\Phi((u-v)/2)\). For \(u,-v\), \[\Phi(u)+\Phi(-v)=2\Phi\left(\frac{u-v}{2}\right)\Phi\left(\frac{u+v}{2}\right).\] The right sides are equal, so \(\Phi(u)+\Phi(v)=\Phi(u)+\Phi(-v)\). Hence \(\Phi(v)=\Phi(-v)\) for all \(v\).
A. Source analysis. Main objects: a function with mean-type arguments (u+v)/2 and (u-v)/2.
B. Insufficient first move. Direct substitution u=0 does not immediately prove evenness.
C. Hidden observation. Replacing v by -v swaps the two factors on the right.
D. Required move. One must compare two equations rather than solve the equation completely.
E. Number of ideas. One key idea: symmetric variable replacement.
F. Difficulty justification. Regional level 6: the solution is short, but the substitution is hidden among two means.
G. Why it is not one-step. It is not one-step: a second equation must be built and compared.