Algebra and little number theory

To prove that $f(x) = x+1$ __(1) for all rational numbers , we start to prove for the basic step that $f(x) = x+1$ satisfies for all natural numbers, then integers, then positive rational numbers. After that, we combine both negative integers, and positive rational numbers to get all negative rational numbers.

It's easy to prove that (1) is true for all positive integers.

Prove for all integers

Plugging $x,y \rightarrow 0$ to (1) we get $f(0) = f(0)^{2} - f(0) + 1$

which gives $f(0) = 1$

Plugging $x\rightarrow 1,y \rightarrow -1$ to (1) we get $f(-1) = f(1)f(-1) - f(0) + 1$

$f(-1) = 2f(-1)$

Which gives $f(-1) = 0$

Note: $m,n$ are natural numbers.

Plugging $x \rightarrow -1, y \rightarrow n$ to (1)

$f(-n) = -f(n-1) + 1$

If $n = 1$ gives $f(-1) = 0$ which is the same to $f(-n) = -n+1$ .

Therefore, we proved that $f(x) = x+1$ for all integers $x$ . (A)

Prove for all rational numbers (quite hard though)

Plugging $x\rightarrow n, y\rightarrow \displaystyle \frac{1}{n}$ to (1)

$f(1) = f(n)f(\frac{1}{n}) - f(n+\frac{1}{n} + 1)$ (2)

Plugging $x = 1, y = m+\displaystyle \frac{1}{n}$ we get

$f(m+\frac{1}{n}) = 2f(m+\frac{1}{n}) - f(m+\frac{1}{n} + 1) + 1$

Which is

$f(m+\frac{1}{n} + 1) = f(m+\frac{1}{n}) + 1$ (3)

By induction on $m$ , we get that $f(m+\frac{1}{n}+1) = m+f(\frac{1}{n} + 1)$ .

Which can be reduced to $f(m+\frac{1}{n}) = m+f(\frac{1}{n})$

Plugging $m\rightarrow n$ from (3) into (2) we get

$1 = (n+1)f(\frac{1}{n}) - (n+1+f(\frac{1}{n}))+1$

Which gives $f(\frac{1}{n}) = \frac{n+1}{n} = \frac{1}{n} + 1$

Plugging $x = m, y = \frac{1}{n}$ we get

$f(\frac{m}{n}) = f(m)f(\frac{1}{n}) - f(m+\frac{1}{n}) + 1$

$f(\frac{m}{n}) = (m+1)(\frac{1}{n} + 1) - (m+\frac{1}{n}+1)+1$

Which simplifies to $f(\frac{m}{n}) = \frac{m}{n} + 1$

Therefore, $f(x) = x+1$ for all positive rational numbers. (B)

Prove for negative rational numbers

Note: $q$ is rational numbers.

Plugging $x \rightarrow -1, y \rightarrow q$ we get

$f(-q) = f(q)f(-1) - f(q-1) + 1$

Combine both (A) and (B) we get

$f(-q) = -q + 1$

Therefore, $f(x) = x+1$ for all rational numbers. ~~~

My fingers hurt. XD

Plugging in $y = 1$ to condition 2, we see that $f(x+1) = f(x) + 1$ for all $x$ . From condition 1, this implies that $f(x) = x+1$ for all positive integers $x$ . So $f(566) = 567$ , which is $3$ mod $6$ , so the answer is $\fbox{81}$ .