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Find all functions $f : \mathbb{N} \to \mathbb{N}$ satisfying $$f(m)[(f(n))^2-1]=f(n)[f(m+n)-f(m-n)]$$ $\forall m>n$.

Attempted work :

Let $P(m,n)$ denote $f(m)[(f(n))^2-1]=f(n)[f(m+n)-f(m-n)], \forall m>n$.

Case 1 : $f(n)=1, \forall n \in \mathbb{N}$.

Case 2 : $\exists k \in \mathbb{N}, f(k)>1$

$P(m,k) : f(m)[(f(k))^2-1]=f(k)[f(m+k)-f(m-k)]$

Since $ \text{gcd} (f(k), f(k)^2-1) = 1$ so $f(k) \mid f(m), \forall m \in \mathbb{N}, m>k$

I don't know how to proceed.

user403160
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  • Do you consider $0$ to be an element of $\mathbb{N}$? – Mathematician 42 Oct 13 '17 at 07:44
  • Taking $m=1$ and $n=0$, we get that $$f(1)(f(0)^2-1)=f(0)(f(1)-f(1))=0.$$ Thus $f(1)(f(0)^2-1)=0$. Thus either $f(1)=0$ or $f(0)=1$. – Mathematician 42 Oct 13 '17 at 07:46
  • @ Mathematician 42. Thank you. $\mathbb{N}$ is positive integers. – user403160 Oct 13 '17 at 07:48
  • Maybe I should ask whether the people making this question considered $0$ to be a natural number. That makes a huge difference for these kind of questions. Anyway, if you allow zero I can solve this one, without that I'm not sure how to do this. – Mathematician 42 Oct 13 '17 at 07:51
  • The word positive integer is still ambiguous. Some people consider positive to be $n>0$, others say it's $n\geq 0$. For some reason the world is divided in two groups on this issue. – Mathematician 42 Oct 13 '17 at 07:53
  • If $0$ is included, we say $\mathbb{N} \cup {0}$. – user403160 Oct 13 '17 at 08:01
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    @carat It depends, in many European countries the notation $\mathbb N$ contains $0$ and else we write $\mathbb N^*$ while in USA it does not include $0$ it seems. – zwim Oct 13 '17 at 08:05
  • Anyway, where does this equation comes from? If it is reverse problem, maybe a closed form would help tackle this problem. Seems, $a^n$ would work. – zwim Oct 13 '17 at 08:11
  • @zwim. Thank you for the information, we use US notation. It's a contest problem. I don't know where the equation comes from. – user403160 Oct 13 '17 at 08:14

1 Answers1

2

When $f(n)=a^n$ then $f$ verifies the functional equation.

$f(n)(f(m+n)-f(m-n))=a^n(a^{m+n}-a^{m-n})=a^m(a^{2n}-1)=f(m)(f(n)^2-1)$.


So let's set $f(1)=a\in\mathbb N^*$ and see if we can bring something by induction on $f(n)=a^n$.

$f(1)(f(m+1)-f(m-1))=f(m)(f(1)^2-1)\\\implies a(f(m+1)-a^{m-1})=a^m(a^2-1)\\\implies af(m+1)=a^{m+2}-a^m+a^m=a^{m+2}\\\implies f(m+1)=a^{m+1}$

Unfortunately for the induction to work we need both $f(m)$ and $f(m-1)$ to be verified.

Thus we need $f(1)=a$ and $f(2)=a^2$ at initial step.

But we know nothing about $f(2)$


From now on, I'm using a CAS, because it is tedious to calculate everything

  • $F_1=f(1)=a$

  • $F_n=F_{n-2}+F_{n-1}(a^2-1)/a$

The values $F_n=f(n)$ obtained through formula $$f(1)(f(m+1)-f(m-1))=f(m)(f(1)^2-1)$$

  • $B_n=F_{n-2}(F_2^2-1)/F_2+F_{n-4}$

The values $B_n=f(n)$ obtained through formula $$f(2)(f(m+2)-f(m-2))=f(m)(f(2)^2-1)$$

Let's set $F_2=b$

$F_3=a+ab-\frac ba$

Since $F_3$ integer then $a\mid b$

Let's set $F_2=ab$

$F_3=a+a^2b-b$

$F_4=-ab+a^2-1+a^3b+\frac ba$

Since $F_4$ integer then $a\mid b$

Let's set $F_2=a^2b$

$F_3=a+a^3b-ab$

$F_4=-a^2b+a^2-1+a^4b+b$

$F_5=-a-a^3b+ab+a^3+a^5b+\frac 1a-\frac ba$

Since $F_5$ integer then $a\mid (b-1)$

Let's set $F_2=a^2(1+ka)\quad$ with $k\ge 0$

patience, we are getting to it...

$F_3=a^3+ka^4-ka^2$

$F_4=-ka^3+a^4+a^5k+ka$

$F_5=-ka^4+ka^2+a^5+ka^6-k$

$F_6=ka^3-a^5k-ka+a^6+a^7k+\frac ka$

$B_5=\dfrac{(a^5+3ka^6+3a^7k^2+a^8k^3-ka^4-2a^5k^2-a^6k^3+k)}{(1+ka)}$

$B_6=\dfrac{-ka^6-2a^7k^2-a^8k^3+ka^2+a^7+3a^8k+3k^2a^9+a^{10}k^3+3a^5k^2+a^6k^3-k+2ka^4}{a(1+ka)}$

Now by definition $F_5=B_5$ and $F_6=B_6$

Unfortunately, the expressions are complicated, but I found that

$(F_5-B_5)+a(F_6-B_6)=\dfrac{-ka^2\overbrace{(ka^3+a^2+1)}^{>0}\overbrace{(a^5k+a^4+1)}^{>0}}{1+ka}=0$

Thus $k=0$ is forced.

Consequently, $f(2)=a^2$ and the induction base hypothesis is verified.

Finally $f(n)=a^n$ for $a\in\mathbb N^*$ are the unique solutions of the equation.

zwim
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  • Note: from $F_n=F_{n−2}+F_{n−1}(a^2−1)/a$ we can deduce directly $F_n=\alpha a^n+\beta\frac{(-1)^n}{a^n}$. Although I don't see how to progress further not knowing $F_2$. Can we simply dismiss $a\neq1,\beta\neq0$ and $F_n=\alpha+\beta(-1)^n$ otherwise ? It doesn't seems to be that straightforward, maybe I'm overviewing something. – zwim Oct 13 '17 at 10:40
  • What is the idea behind the steps following the definition for $B_n$? I was able to follow your proof, but I'm not clear on the reason for them –  Oct 13 '17 at 17:23
  • Formula for $m,n=1$ is not enough to reach a conclusion (see my comment above). So we have to exploit the formula for $m,n=2$ too. The first couples with $m>n$ are $m=3,n=2$ leading to $f(5)$ and $m=4,n=2$ leading to $f(6)$. I give them a different name $B_5$, $B_6$ but it is only for clarity. Solving $F5=B5=f(5)$ does not force $k=0$ so I was forced to use $F6=B6=f(6)$ condition too. I know this is complicated but I don't know how to shorten this proof. For instance to dismiss the solution $f(n)=\alpha+\beta(-1)^n$ I checked that I need also to compute $f(5)$ and $f(6)$ to get rid of $\beta$. – zwim Oct 13 '17 at 17:34
  • Thanks for the response. Actually, though, I meant to ask about your setting different values for $f(2)$ - $b,ab$ and so on. What was the idea there? –  Oct 13 '17 at 17:44
  • Regarding $f(n)=\alpha+\beta(-1)^n$. So that was another case you tried? –  Oct 13 '17 at 17:45
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    I first set $f(2)=b$ most general I can do. Then I reach conclusion $a$ divides $b$ so $b=ab'$ and $f(2)=ab'$. Idem $a$ divides $b'$ so $f(2)=a^2b''$. Since $b,b',b''$ are dummy, I just called them all $b$. Next $a$ divides $(b''-1)$ so $b''=1+ka$ so I end up with $f(2)=a^2(1+ka)$. – zwim Oct 13 '17 at 17:53
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    For the other question, see my first comment. This is the case $a=1$. – zwim Oct 13 '17 at 17:54
  • Thanks for your patience and help! –  Oct 13 '17 at 18:05
  • Thank you very much, zwim. I understand your answer. I could find $F_2$ by hand. In the test, we cannot use calculating equipment and time allowed is 20 minutes. I think I can't do it in time :) – user403160 Oct 14 '17 at 03:24