IMO 2012 Shortlist A5

IMO 2012 Shortlist A5

Category: Algebra

Problem

Find all functions f : R → R that satisfy the conditions f(1 + xy) − f(x + y) = f(x)f(y) for all x,y ∈ R and f(−1) 6= 0. Solution. The only solution is the function f(x) = x − 1, x ∈ R. We set g(x) = f(x)+1 and show that g(x) = x for all real x. The conditions take the form g(1 + xy) − g(x + y) = g(x) − 1
g(y) − 1
for all x,y ∈ R and g(−1) 6= 1. (1) Denote C = g(−1) − 1 6= 0. Setting y = −1 in (1) gives g(1 − x) − g(x − 1) = C(g(x) − 1). (2) Set x = 1 in (2) to obtain C(g(1) − 1) = 0. Hence g(1) = 1 as C 6= 0. Now plugging in x = 0 and x = 2 yields g(0) = 0 and g(2) = 2 respectively. We pass on to the key observations g(x) + g(2 − x) = 2 for all x ∈ R, (3) g(x + 2) − g(x) = 2 for all x ∈ R. (4) Replace x by 1 − x in (2), then change x to −x in the resulting equation. We obtain the relations g(x)−g(−x) = C(g(1−x)−1), g(−x)−g(x) = C(g(1+x) −1). Then adding them up leads to C(g(1 − x) + g(1 + x) − 2) = 0. Thus C 6= 0 implies (3). Let u,v be such that u + v = 1. Apply (1) to the pairs (u,v) and (2 − u,2 − v): g(1 + uv) − g(1) = g(u) − 1
g(v) − 1
, g(3 + uv) − g(3) = g(2 − u) − 1
g(2 − v) − 1
. Observe that the last two equations have equal right-hand sides by (3). Hence u+v = 1 implies g(uv + 3) − g(uv + 1) = g(3) − g(1). Each x ≤ 5/4 is expressible in the form x = uv + 1 with u + v = 1 (the quadratic function t2 −t+(x−1) has real roots for x ≤ 5/4). Hence g(x+2)−g(x) = g(3)−g(1) whenever x ≤ 5/4. Because g(x) = x holds for x = 0,1,2, setting x = 0 yields g(3) = 3. This proves (4) for x ≤ 5/4. If x > 5/4 then −x < 5/4 and so g(2 − x) − g(−x) = 2 by the above. On the other hand (3) gives g(x) = 2−g(2−x), g(x+2) = 2−g(−x), so that g(x+2)−g(x) = g(2−x)−g(−x) = 2. Thus (4) is true for all x ∈ R. Now replace x by −x in (3) to obtain g(−x) + g(2 + x) = 2. In view of (4) this leads to g(x) + g(−x) = 0, i. e. g(−x) = −g(x) for all x. Taking this into account, we apply (1) to the pairs (−x,y) and (x,−y): g(1 − xy) − g(−x + y) = g(x) + 1
1 − g(y)
, g(1 − xy) − g(x − y) = 1 − g(x)
g(y) + 1
. Adding up yields g(1 − xy) = 1 − g(x)g(y). Then g(1 + xy) = 1 + g(x)g(y) by (3). Now the original equation (1) takes the form g(x + y) = g(x) + g(y). Hence g is additive. By additvity g(1 + xy) = g(1) + g(xy) = 1 + g(xy); since g(1 + xy) = 1 + g(x)g(y) was shown above, we also have g(xy) = g(x)g(y) (g is multiplicative). In particular y = x gives g(x2 ) = g(x)2 ≥ 0 for all x, meaning that g(x) ≥ 0 for x ≥ 0. Since g is additive and bounded from below on [0,+∞), it is linear; more exactly g(x) = g(1)x = x for all x ∈ R. In summary f(x) = x − 1, x ∈ R. It is straightforward that this function satisfies the requirements. Comment. There are functions that satisfy the given equation but vanish at −1, for instance the constant function 0 and f(x) = x2 − 1, x ∈ R.16