Problem: The function f satisfies the functional equation
f(x)+f(y)=f(x+y)βxyβ1
for every pair x,y of real numbers. If f(1)=1, then the number of integers nξ =1 for which f(n)=n is
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. infinite
Solution:
Substitute x=1 into the functional equation and solve fion the first term on the right sode to ohtain f(y+1)=f(y)+y+2. Sinie f(1)=1, one sees by successively substituling y=2,3,4,β¦ thal f(y)>0 for every positive integer. Therefine, for y a pusitive integer. f(y+1)>y+2>y+1. and f(n)=n has no solutions for integers n>1. Soiving the abuve equation for f(y) yields
f(y)=f(y+1)β(y+2)
Successively substituting y=0,β1,β2,β¦ into this equation yields f(0)=β1,f(β1)=β2,f(β2)=β2,f(β3)=β1.f(β4)=1. Now f(β4)>0 and. for y<β4.β(y+2)>0. Thus, for y<β4,f(y)>0. Therefore, f(n)ξ =n for n<β4; and the solutions n=1,β2 are the only ones.