Find all $f:Z→Z$ which satisfy $f(m+n)(f(m)−f(n))=f(m−n)(f(m)+f(n))$

Question

question -

Find all $f:Z→Z$ which satisfy the conditions $f(1)=1$ and $f(m+n)(f(m)−f(n))=f(m−n)(f(m)+f(n))$ for all integers $m,n$

my try -

I first Prove that $f(0)=0$ and hence $f(n)(f(n)+f(−n))=0$ for all $n∈Z$. Changing $n$ to $−n$, i conclude that $f(n)+ f(−n)=0$. so hence i get that f is odd function.....

Taking $m=2$ and $n=1$, I conclude that $(f(3)−1)(f(2)−1)=2$.

This gives me four possibilities:

(a)$f(2)=2$, $f(3)=3$;

(b)$f(2)=3$, $f(3)=2$ ;

(c)$f(2)=0$, $f(3)=−1$

and (d) $f(2)=−1$,$f(3)=0$.

I Showed that (a) leads to $f(n)=n$ for all n and (b) is not possible by checking some values .....but i am not able to find $f$ in cases (c) and (d).....

any help will be appreciated

thankyou

Answer 1

Case c)

Taking $n=1$ we get $$\begin{align} f(2+1) (f(2)-f(1)) &= f(2-1) (f(2)+f(1)) \implies f(3) = -1 \\ f(3+1) (f(3)-f(1)) &= f(3-1) (f(3)+f(1)) \implies f(4) = 0 \\ f(4+1) (f(4)-f(1)) &= f(4-1) (f(4)+f(1)) \implies f(5) = 1 \\ \vdots \\ f(2k) = 0,\ &f(2k+1) = (-1)^k \end{align}$$ This result can even be written as one formula: $f(n) = \sin\frac{n\pi}{2}.$

Case d should now be easy.

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Induction proof

We know that $f(0) = 0$, $f(1) = 1,$ $f(2) = 0,$ $f(3) = -1.$

Now suppose that for some $k \in \mathbb{Z}$ we have $f(4k+0) = 0,$ $f(4k+1) = 1,$ $f(4k+2) = 0,$ $f(4k+3) = -1.$ Then, $$ f((4k+3)+1) (f(4k+3) - f(1)) = f((4k+3)-1) (f(4k+3) + f(1)) \\ f(4k+4) (-1 - 1) = 0 (-1 + 1) \\ \therefore f(4(k+1)+0) = 0. \\ $$ $$ f((4k+4)+1) (f(4k+4) - f(1)) = f((4k+4)-1) (f(4k+4) + f(1)) \\ f(4k+5) (0 - 1) = (-1) (0 + 1) \\ \therefore f(4(k+1)+1) = 1. \\ $$ $$ f((4k+5)+1) (f(4k+5) - f(1)) = f((4k+5)-1) (f(4k+5) + f(1)) \\ f(4k+6) (-1 - 1) = 0 (-1 + 1) \\ \therefore f(4(k+1)+2) = 0. \\ $$ $$ f((4k+6)+1) (f(4k+6) - f(1)) = f((4k+6)-1) (f(4k+6) + f(1)) \\ f(4k+7) (0 - 1) = 1 (0 + 1) \\ \therefore f(4(k+1)+3) = -1. \\ $$

Since the assertions $f(4k+0) = 0,$ $f(4k+1) = 1,$ $f(4k+2) = 0,$ $f(4k+3) = -1$ are valid for $k=0$ and the validity for any $k$ implies validity for $k+1$ we get that they are valid for all $k \in \mathbb{N}_0.$

How about negative $k$? We can use that $f$ is odd. Let $k<0.$ Then $-k\geq 0$ so $$ f(4k+0) = -f(-(4k+0)) = -f(4(-k-1)+4) = -0 = 0, \\ f(4k+1) = -f(-(4k+1)) = -f(4(-k-1)+3) = -(-1) = 1, \\ f(4k+2) = -f(-(4k+2)) = -f(4(-k-1)+2) = -0 = 0, \\ f(4k+3) = -f(-(4k+3)) = -f(4(-k-1)+1) = -1. $$ Thus the asserted formulas are also valid for $k<0.$