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$f:\Bbb Z\to\Bbb Z$ is such that $f(11)=1$ and $f(a)f(b)=f(a+b)+f(a-b)$ for all integers $a,b$. What is $f(550)$?

I've first set $f(11)f(0)=2f(11)$, which yields $f(0)=0,2$. Yet I know that 0 won't work because that was a hint. I've also heard that there was a pattern.

The answer is supposed to be $-1$, but I don't know how to get there.

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Notice that

$$f(11n)f(11) = f(11(n +1)) + f(11(n-1))$$

and since $f(11) = 1$,

$$f(11n) = f(11(n+1)) + f(11(n-1))$$

Define $a_n = f(11n)$ and we can set up a recurrence relation

$$a_n = a_{n+1} + a_{n-1}$$

or

$$a_{n+1} = a_n - a_{n-1}$$

which resolves periodically to

$$a_{n+3} = - a_n\implies a_{n+6} = a_n$$

Now, note that

$$\begin{align}a_{50} &= a_{8\cdot6 + 2}\\&=a_2\end{align}$$

Then, $$f(11)f(11) = f(22) + f(0)$$ $$1 = f(22) +2$$ $$f(22) = -1$$ so $$a_{50} = a_{2} = f(22) = -1$$

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  • $\begingroup$ $f(550)$ equals $a_{50}$, not $a_{55}$. $\endgroup$ – Mario Carneiro Sep 8 '16 at 3:13
  • $\begingroup$ @MarioCarneiro Fixed $\endgroup$ – Yiyuan Lee Sep 8 '16 at 6:32
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We start with $f(11n)f(11)=f(11(n+1))+f(11(n−1))$ for every integer $n$.

and since $f(11)=1$,

$f(11n)=f(11(n+1))+f(11(n−1))$. Let $a_{n} = f(11n)$ and we have the following relation:

$a_{n+1} = a_{n} - a_{n-1}$

From this we get $a_{n+6} = a_{n}$.

Based on this last relation, we can write $f(550)=a_{50} = a_{6.8+2} = a_{2} = f(22)$

But, $f(11)f(11) = f(22)+f(0)$ and then we get $f(22) = 1 - 2 = -1$.

We conclude that $f(550) = -1$. As required!

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  • 2
    $\begingroup$ Don't refer to other answers with "above" or "below", because the SE platform reorders the answers in unpredictable ways. $\endgroup$ – Mario Carneiro Sep 8 '16 at 3:20
  • $\begingroup$ Of course. I just couldn't coment your trivial mistake because I was not able to coment yet, since my reputation is low. $\endgroup$ – C.E.A.H - AMK963 Sep 8 '16 at 9:48

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