XXXII - I - Problem 10

Determine all functions $ f $ mapping the set of all rational numbers $ \mathbb{Q} $ to itself that satisfy the following conditions:
a) $ f(1)=2 $,
b) $ f(xy) = f(x)f(y)-f(x+y)+1 $ for $ x, y \in \mathbb{Q} $.
If the function $ f $ satisfies conditions a) and b), then the function $ g: \mathbb{Q} \to \mathbb{Q} $ defined by the formula $ g(x) = f(x)- 1 $ satisfies the conditions
a')  $ g(1) = 1 $
and $ g(xy)+1 = (g(x)+1)(g(y)+1)-g(x+y)-1+1 $, $ g(xy)+1 = g(x)g(y)+g(x)+g(y)+1-g(x+y) $, i.e.
b')  $ g(xy)+g(x+y) = g(x)g(y)+g(x)+g(y) $.
Substituting $ y = 1 $ in the last equality, we get






thus, by a')






Hence






Substituting $ x = 0 $ and $ x = -1 $ in the last equality, we will determine, remembering a')












By obvious inductive reasoning, it follows from this and (*), that






for any integer $ n $, and






for an integer $ m $ and a rational $ x $.
Now let $ m $ be an integer, and $ n $ a natural number.
To determine $ g\left( \frac{m}{n} \right) $, substitute $ x = m $, $ y = \frac{1}{n} $ in equality b).




Since $ g(m)= m $ and $ g\left( m+\frac{1}{n} \right)=m+g \left( \frac{1}{n} \right) $, we have












If $ m = n $, then $ 1 = g(1) = n \cdot g \left( \frac{1}{n} \right) $, from which it follows that






Therefore






This means that if the function $ g $ satisfies conditions a') and b'), then $ g(x) = x $ for every rational $ x $. Therefore, if the function $ f $ satisfies conditions a) and b), then $ f(x) = x+ 1 $. On the other hand, the function $ f $ defined by this formula obviously satisfies conditions a) and b). The only solution to the problem is therefore the function $ f: \mathbb{Q} \to \mathbb{Q} $, defined by the formula $ f(x) = x+1 $.