Finding $f(9999)$ if $f(1) = 2$ and $f(mn) = f(m) \, f(n) - f(m+n) + 1001$ for $m$ or $n$ equal to $1$

Question

Given a function $f$ is defined for integers $m$ and $n$ as given: $$f(mn) = f(m)\,f(n) - f(m+n) + 1001$$ where either $m$ or $n$ is equal to $1$, and $f(1) = 2$.

The problem itself is to prove that $$f(x) = f(x-1) + 1001$$ in order to find the value of $f(9999)$.

As such, what I've already tried is:
Replacing $n$ as $1$ and $m$ as $x$, and trying to solve from there.

$$f(x) = f(1) * f(x) - f(x+1) + 1001$$ $$f(x) = 2 * f(x) - f(x+1) + 1001,$$

Although I'm not sure how I should continue from here, it did occur that the negative sign could be manipulated in some way, but I'm fairly certain that $-f(x+1)$, cannot be rewritten as $f(-x-1)$, or anything of the sorts.

So if anyone has any thoughts or insight as how I should proceed, it would be greatly appreciated.

Answer 1

You got $$f(x)=2f(x)-f(x+1)+1001$$Now, let's find $f(x+1)$ using $f(x)$:$$f(x)=2f(x)-f(x+1)+1001\iff f(x+1)=f(x)+1001$$Now, let $y=x+1, x=y-1$ you get $$f(y)=f(y-1)+1001$$

(Note, both $x$ and $y$ are just names of the variables, so if it is more convenient to you, you can rewrite the final line using $x$ instead of $y$)

Answer 2

Given a function f is defined for integers m and n as given: f(mn)=f(m)f(n)−f(m+n)+1001 where either m or n is equal to 1

As multiplication and addition is commutative we can assume, wolog $n = 1$ and that is just a convoluted way of simply saying $f(m) = 2f(m) -f(m+1) +1001$ or $f(m+1) = f(m) + 1001$. Which means, by induction, $f(m) = f(1) + 1001(m-1)$.

And as $f(1) = 2$ we have $f(m) = 2 + 1001(m-1)$ for all $m\in \mathbb Z; m \ne 1$.

So $f(9999) = 2 + 1001*9998=10008000$

Answer 3

Disclaimer. This is an attempt to see what happens if I drop the conditions that $f(1)=2$ and that at least one of $m$ and $n$ in the functional equation must be $1$. That is, this answer does not answer the OP's question, but I think it is a nice relevant question. Since the work is quite long, I have to put it here as an answer.

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Fix $k\in\mathbb{C}$. Let $f:\mathbb{Z}_{>0}\to\mathbb{C}$ be a function satisfying $$f(mn)=f(m)\,f(n)-f(m+n)+k$$ for all $m,n\in\mathbb{Z}_{>0}$. Write $a:=f(1)$. Then, plugging in $m=1$, we have $$f(n+1)=(a-1)\,f(n)+k\tag{*}$$ for all $n\in\mathbb{Z}_{>0}$. We have three notable scenarios: $a=1$, $a=2$, and $a\notin\{1,2\}$.

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If $a=1$, then $f(n)=k$ for all $n\in\mathbb{Z}_{>1}$. That means $$k=f(4)=f(2\cdot2)=f(2)\,f(2)-f(2+2)+k=k\cdot k-k+k\,.$$ Thus, $k^2=k$, or $k\in\{0,1\}$. Hence, there are two possible solutions with $f(1)=1$:

• $k=0$, which gives $f(1)=1$ and $f(n)=0$ for all $n\in\mathbb{Z}_{>1}$;
• $k=1$, which gives $f(n)=1$ for all $n\in\mathbb{Z}_{>0}$.

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Now, we suppose that $a=2$. Thus, $f(n+1)=f(n)+k$ for all $n\in\mathbb{Z}_{>0}$. This leads to $f(n)=2+k(n-1)$ for all $n\in\mathbb{Z}_{>0}$. Consequently, $$3k+2=f(4)=f(2\cdot 2)=f(2)\,f(2)-f(2+2)+k=(k+2)^2-(3k+2)+k\,,$$ yielding $$k^2-k=0\,.$$ Thus, $k=0$ or $k=1$. If $k=0$, then $f(n)=2$ for all $n\in\mathbb{Z}_{>0}$. If $k=1$, then $f(n)=n+1$ for all $n\in\mathbb{Z}_{>0}$.

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Finally, we tackle the case $a\notin \{1,2\}$. Then, the solution to the recurrence relation (*) is $$f(n)=(a-1)^{n-1}\left(a+\frac{k}{a-2}\right)-\frac{k}{a-2}\,.$$ In particular, $f(2)=a(a-1)+k$ and $f(4)=a(a-1)^3+\left((a-1)^2+(a-1)+1\right)k$. Using the same equation $2\,f(4)=\big(f(2)\big)^2+k$ from before, we get $$2a(a-1)^3+2\left((a-1)^2+(a-1)+1\right)k=\big(a(a-1)+k\big)^2+k\,.$$ Thus, $$a(a-2)(a-1)^2=k(k-1)\,.$$ This means $k=(a-1)^2$ or $k=-a(a-2)$. For the subcase $k=-a(a-2)$, we see that $$f(n)=a\text{ for all }n\in\mathbb{Z}_{>0}\,.$$

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For the subcase $k=(a-1)^2$, we get $$f(n)=a(a-1)^{n-1}+(a-1)^2\left(\frac{(a-1)^{n-1}-1}{a-2}\right)\text{ for each integer }n>0\,.$$ For simplicity, write $b:=a-1$, so $k=b^2$ and $f(n)=(b+1)b^{n-1}+b^2\left(\dfrac{b^{n-1}-1}{b-1}\right)$. We have $$f(2)=(b+1)b+b^2\,,\,\,f(3)=2(b+1)b^2\,,$$ $$f(5)=(b+1)b^4+b^2(b^3+b^2+b+1)\,,$$ and $$f(6)=(b+1)b^5+b^2(b^4+b^3+b^2+b+1)\,.$$ Since $f(6)=f(2\cdot3)=f(2)\,f(3)-f(2+3)+k=f(2)\,f(3)-f(5)+b^2$, we obtain $$2b^6+2b^5+b^4+b^3+b^2=(2b^2+b)(2b^3+2b^2)-(2b^5+2b^4+b^3+b^2)+b^2\,.$$ That is, $$b^2(b^2-1)(2b^2-1)=2b^6-3b^4+b^2=0\,.$$ Hence, $b=0$, $b=+1$, $b=-1$, $b=+\dfrac1{\sqrt{2}}$, or $b=-\dfrac1{\sqrt{2}}$.

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Recall that $a\notin\{1,2\}$. If $b=0$, then $a=1$, which is a contradiction. If $b=+1$, then $a=2$, another contradiction. If $b=-1$, then $a=0$, $k=1$, and $$f(n)=\frac{1+(-1)^{n}}{2}=\left\{\begin{array}{ll}0\,,&\text{if }n\text{ is odd}\,,\\ 1\,,&\text{if }n\text{ is even}\,. \end{array}\right.$$

For $b=+\dfrac1{\sqrt{2}}$, we get $a=1+\dfrac1{\sqrt{2}}$, $k=\dfrac{1}{2}$, and $$f(n)=1+\frac1{\sqrt{2}}\text{ for all positive integers }n\,.$$ For $b=-\dfrac1{\sqrt{2}}$, we get $a=1-\dfrac1{\sqrt{2}}$, $k=\dfrac12$ and $$f(n)=1-\frac1{\sqrt{2}}\text{ for all positive integers }n\,.$$

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From the work above, we can now conclude our proposition. In particular, for $k=1001$, the proposition states that there are two solutions: $$f(n)=1+10\sqrt{10}\text{i}\text{ for each }n\in\mathbb{Z}_{>0}$$ and $$f(n)=1-10\sqrt{10}\text{i}\text{ for each }n\in\mathbb{Z}_{>0}\,.$$ That is, if you allow $m$ and $n$ in the functional equation to be any positive integer and drop the condition that $f(1)=2$, then the solutions look wildly different.

Proposition: Let $k$ be a complex number. Then, all solutions $f:\mathbb{Z}_{>0}\to\mathbb{C}$ to the functional equation $$f(mn)=f(m)\,f(n)-f(m+n)+k\text{ for all positive integers }m\text{ and }n$$ is given by the list below.
(a) If $k=-a(a-2)$ for some $a\in\mathbb{C}$, then $$f(n)=a\text{ for all }n\in\mathbb{Z}_{>0}\,.$$ There are two values of $a$ for all $k\in\mathbb{C}\setminus\{1\}$, namely, $a=1-\sqrt{1-k}$ and $a=1+\sqrt{1-k}$. For $k=1$, there is only one value of $a$, i.e., $a=1$.
(b) In the case $k=0$, there is one more solution apart from two solutions covered in (a), and this extra solution is given by $$f(n)=\left\{\begin{array}{ll}1\,,&\text{if }n=1\,,\\0\,,&\text{for }n=2,3,4,\ldots\,.\end{array}\right.$$
(c) In the case $k=1$, there are two more solutions apart from the one solution covered in (a). These solutions are the linear function $$f(n)=n+1\text{ for all }n\in\mathbb{Z}_{>0}$$ and the alternating function $$f(n)=1-(n\,\text{mod}\,2)=\left\{\begin{array}{ll}0\,,&\text{if }n\text{ is odd}\,,\\ 1\,,&\text{if }n\text{ is even}\,. \end{array}\right.$$