MathLabs

Problem 5

Find all functions f:R+→R+f:\mathbb{R}^+\to\mathbb{R}^+ such that (z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x)(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x) for all positive real numbers x,y,zx,y,z.
Step 7 of 8: Pin down the additive constant
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f(x+y)=f(x)+f(y)+b,b=f(2)−2f(1)=f(2)−2,4+2b=2f(2)=f(f(2)+f(2))=f(f(2))+f(f(2))+b=4+b  ⟹  b=0f(x+y)=f(x)+f(y)+b,\quad b=f(2)-2f(1)=f(2)-2,\quad 4+2b=2f(2)=f(f(2)+f(2))=f(f(2))+f(f(2))+b=4+b \implies b=0
Detailed analysis

Applying additivity over equal sums twice, f(x+y)+f(1)=f(x)+f(y+1)f(x+y)+f(1)=f(x)+f(y+1) and f(y+1)+f(1)=f(y)+f(2)f(y+1)+f(1)=f(y)+f(2); combining gives f(x+y)=f(x)+f(y)+bf(x+y)=f(x)+f(y)+b for all x,y>0x,y>0, where b=f(2)−2f(1)=f(2)−2b=f(2)-2f(1)=f(2)-2. Using yf(x)=f(xf(y))yf(x)=f(xf(y)), the involution f(f(y))=yf(f(y))=y, and this additive relation at x=y=2x=y=2, one computes 4+2b=2f(2)=f(2f(2))=f(f(2)+f(2))=f(f(2))+f(f(2))+b=4+b4+2b=2f(2)=f(2f(2))=f(f(2)+f(2))=f(f(2))+f(f(2))+b=4+b, forcing b=0b=0.