x=y=0 ==> f(0)=f(0)² ==> f(0)=0 ou f(0)=1
Si f(0)=0 ==> f(x)=0 qqs x
On suppose alors f(0)=1
qqs y , f(y)+f(-y)=2f(y) ==> qqs y , f(y)=f(-y) ==> f paire
f(c)=0 , qqs x ,
f(x+4c)+f(x+2c)=2f(x+3c)f(c)=0
f(x+2c)+f(x)=2f(x+c)f(c)=0
==> f(x+4c)=f(x)
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