Find all strictly increasing functions f:N->N such that (f(x) + f(y))/(1 + f(x + y)) is a non-zero natural number, for all x, y∈N.
f(x) and f(y) are positive, so 0 < (f(x) + f(y))/(1 + f(x+y))
x and y are positive so x+y is larger than x and y individually, and since f() is strictly increasing then f(x+y) is greater than f(x) and f(y) individually.
Then (f(x) + f(y))/(1 + f(x + y)) < 2*(f(x+y))/(1 + f(x+y))
Decreasing the denominator by 1 will increase the value of the fraction, then 2*(f(x+y))/(1 + f(x+y)) < 2*(f(x+y))/f(x+y) = 2.
Combining all the inequalities gives 0 < (f(x) + f(y))/(1 + f(x+y)) < 2.
But (f(x) + f(y))/(1 + f(x+y)) must be an integer. The only integer satisfying the inequality is 1, therefore (f(x) + f(y))/(1 + f(x+y)) = 1.
(f(x) + f(y))/(1 + f(x+y)) = 1 can be rearranged into f(x) + f(y) = 1 + f(x+y).
Let x=z+1 and y=z+1. Then 2*f(z+1) = 1 + f(2z+2)
Let x=z and y=z+2. Then f(z) + f(z+2) = 1 + f(2z+2)
Combining the last two equations yields f(z) + f(z+2) = 2*f(z+1).
Rearrange this into f(z+2)-f(z+1) = f(z+1)-f(z). With z, z+1, and z+2 being consecutive integers, this implies that f(z), f(z+1), and f(z+2) are colinear.
Thus f(x) is a linear function. Let f(x) = mx+b. Then f(x) + f(y) = 1 + f(x+y) becomes (mx+b) + (my+b) = 1 + m*(x+y) + b.
All terms with 'm' cancel each other out leaving 2b=b+1, which makes b=1. Therefore f(x) = m*x+1 for positive integers m.
Solution
