Fix a real number a>1, and take a new variable t. For the values f(t),f(t2), f(at) and f(a2t2), the relation (1) provides a system of linear equations:
x=y=t:x=at,y=at:x=a2t,y=t:x=y=at:(t+t1)f(t)(at+ta)f(at)(a2t+a2t1)f(t)(at+at1)f(at)=f(t2)+f(1)=f(t2)+f(a2)=f(a2t2)+f(a21)=f(a2t2)+f(1)
In order to eliminate f(t2), take the difference of (2a) and (2b); from (2c) and (2d) eliminate f(a2t2); then by taking a linear combination, eliminate f(at) as well:
(t+t1)f(t)−(at+ta)f(at)=f(1)−f(a2) and (a2t+a2t1)f(t)−(at+at1)f(at)=f(1/a2)−f(1), so ((at+at1)(t+t1)−(at+ta)(a2t+a2t1))f(t)=(at+at1)(f(1)−f(a2))−(at+ta)(f(1/a2)−f(1)).
Notice that on the left-hand side, the coefficient of f(t) is nonzero and does not depend on t:
(at+at1)(t+t1)−(at+ta)(a2t+a2t1)=a+a1−(a3+a31)<0.
After dividing by this fixed number, we get
f(t)=C1t+tC2
where the numbers C1 and C2 are expressed in terms of a,f(1),f(a2) and f(1/a2), and they do not depend on t. The functions of the form (3) satisfy the equation:
(x+x1)f(y)=(x+x1)(C1y+yC2)=(C1xy+xyC2)+(C1xy+C2yx)=f(xy)+f(xy).