Why does $f(1) = f(0) + 1$? I do not understand this answer. While I understand the outputs for $f(0)$ and $f(1)$, why at the end does $f(1) = f(0) + 1$?  Where does the $f(1)$ come from? Previously he stated $f(0+1) = f(0) + 1$ which is part of the $f(x + 1)$ function, but why does he make the jump next to $f(1)$? See question below:
We have:
$f(x) = ax^2 + bx + c \tag{i}$
$f(x  +1) = f(x) + x + 1 \tag{ii}$
Substituting $x = 0$ in (i):
$f(0) = c \tag{iii}$
Substituting $x = 1$ in (i)
$f(1) = a + b + c \tag{iv}$
Substituting $x = 0$ in (ii)
$$f(0 + 1) = f(0) + 1$$
Therefore, $f(1) = f(0) + 1$.
Therefore $a + b + c = c + 1$, using (iii) and (iv).
Therefore $a + b = 1$.
 A: The train of reasoning is as follows:


*

*Attempt to discover $f(0)$ in two different ways: using the first equation, and using the second equation. The first equation gives us that it is $c$; the second equation expresses it in terms of $f(1)$, so we need to find that next.

*Attempt to determine $f(1)$ using the first equation. The result is $a+b+c$.

*Substitute the now-known value of $f(1)$ (i.e. $a+b+c$) into the expression from the first step.



If you're trying to determine information about an object, a common technique is to try and discover some property of that object from two different angles. For example, 


*

*the answer you quoted finds $f(0)$ from two different angles (using, respectively, the two given equations);

*a combinatorial proof of $\sum_{i=0}^n \binom{n}{i} = 2^n$ finds "the number of subsets of $\{1,2,\dots,n\}$" in two different ways (namely "the number of subsets of size $i$, summed over each $i$" and "the number of ways to specify a subset uniquely").

