Consider the set of all functions f:ℚ→ℚ such that
f(x+f(y))=f(x+y)+f(y) ∀x,y ∈ ℚ.
Find the sum of all possible (distinct) values of f(99).
To find the sum of all possible distinct values of f(99), we need to determine all the possible values of f(99) and then add them up.
Let's begin by substituting x = 99 and y = 0 into the given functional equation:
f(99 + f(0)) = f(99 + 0) + f(0)
f(99 + f(0)) = f(99) + f(0)
Since the domain and codomain of the function are ℚ (rational numbers), let's consider that f(0) = a, where a is a rational number.
Therefore, we now have:
f(99 + a) = f(99) + a
Now, let's substitute x = 99 + a and y = 0 into the functional equation:
f(99 + a + f(0)) = f(99 + a + 0) + f(0)
f(99 + a + f(0)) = f(99 + a) + f(0)
Substituting our previously derived expression, we get:
f(99 + a + a) = f(99 + a) + a
f(99 + 2a) = f(99 + a) + a
We can continue this process by introducing a new variable b:
Let b = 99 + a, so a = b - 99.
Now our equation becomes:
f(b + b - 99) = f(b) + (b - 99)
Simplifying further:
f(2b - 99) = f(b) + (b - 99)
We can repeat this process indefinitely to find a general pattern for f(n):
f(n) = f(n - 99) + (n - 99)
Now, let's analyze the equation we have derived. We see that f(n) can be determined recursively using the equation f(n) = f(n - 99) + (n - 99). This means that the values of f(n) are determined by the values of f(n - 99) for all n.
Starting from f(99), we can recursively calculate the values of f(198), f(297), f(396), and so on.
To summarize, we can obtain the values of f(99) and subsequently calculate the values of f(n) using the recursive equation f(n) = f(n - 99) + (n - 99) for all n.
Once we have obtained all possible distinct values of f(99), we can sum them up to find the final answer.