Find all $f:\mathbb{Z} \to \mathbb{Z}$ satisfying: $\forall m,n\in \mathbb{N}^{+}$,$f(m+n)+f(m-n)=2f(m)f(n)$ Find all $f:\mathbb{Z} \to \mathbb{Z}$ satisfying: $\forall m,n\in \mathbb{N}^{+}$,$f(m+n)+f(m-n)=2f(m)f(n)$ 
I don't know how to solve?
 A: The trivial solution is $f(n)=0$ for all $n$.  Now let's forget about the trivial solution since it's boring.
Your functional equation resembles the cosine sum-to-product formula.  Recall that $$\cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos(\alpha)\cos(\beta).$$
This made me think of Chebyshev polynomials, which are defined by the recursive relation $$T_{m+1}(x)=2T_1(x)T_{m}(x)-T_{m-1}$$ and satisfy $\cos(nx)=T_n(\cos(x))$.  This is exactly what you get when you plug in $n=1$ into your functional equation!  Hence I think the solution is $f(n)=T_n(f(1))$.
A: Applying for $m=n=0$ gives $2f(0)=2f(0)^2$ so $f(0)=0$ or $f(0)=1$.
First case : $f(0)=0$
$f(m)+f(m)=2f(m)f(0)=0\implies f(m)=0$ for $m>0$
$f(n)+f(-n)=2f(0)f(n)=0\implies f(-n)=0$

$f(n)=0, \forall n\in\mathbb Z$

Second case : $f(0)=1$
$f(n)+f(-n)=2f(n)\implies f(-n)=f(n)$
Let's call $a=f(1)$
$f(n+1)+f(n-1)=2f(n)f(1)=2af(n)$
So if we note $u_n=f(n)$ then it satisfies thre relation : $u_{n+2}-2au_{n+1}+a_n=0$
Characteristic equation is $r^2-2ar+1=0$ whose $\Delta=4(a^2-1)$
subcase $a=1$
$r=1$ so $u_n=(bn+c)1^n=bn+c$
$f(0)=c=1$ and $f(1)=b+c=1$ so $b=0$

$f(n)=1,\forall n\in\mathbb Z$ 

subcase $a=-1$
$r=-1$ so $u_n=(bn+c)(-1)^n$
$f(0)=c=1$ and $f(1)=-(b+c)=-1$ so $b=0$

$f(n)=(-1)^n,\forall n\in\mathbb Z$ 

subcase $|a|<1$
Note: since $f(1)=a$ is an integer, then $a=0$ is forced, but let's find it back later... for the fun.
This makes $\Delta<0$ there are two roots $r=a\pm i\sqrt{1-a^2}$
Note that $r\bar r=a^2+1-a^2=1$ so $|r|=1$ and we can associate a trigonometric line $\theta\in[0,\pi]$ with $\cos(\theta)=a$.
$f(n)=b\cos(n\theta)+c\sin(n\theta)$
$f(0)=b=1$ and $f(1)=b\cos(\theta)+c\sin(\theta)=ba+c\sqrt{1-a^2}=a$ so $c=0$.
$f(n)=\cos(n\theta)$
But we require that $f(n)\in\mathbb Z$ so $f(n)$ can be only $\{-1,0,1\}$.
Thus $\theta$ can only be $\{0,\frac{\pi}2,\pi\}$ also $|a|=|\cos(\theta)|<1$ so the only possibility is $\theta=\frac{\pi}2$.

$\displaystyle f(n)=\cos(\frac{n\pi}2)=\frac{i^n+(-i)^n}2,\ \forall n\in\mathbb Z$

subcase $|a|>1$
This makes $\Delta>0$ there are two roots $r=a\pm\sqrt{a^2-1}$
$f(n)=br^n+c\bar r^n$
$f(0)=b+c=1$ and $f(1)=br+c\bar r=(b+c)a+(b-c)\sqrt{a^2-1}=a$ so $b-c=0$.

$\displaystyle f(n)=\frac{r^n+\bar r^n}2$ with $r=a\pm\sqrt{a^2-1}$ and $|a|>1$ integer$

For $u>1$, we can also make the substitution $a=\cosh(u)$ in this case $r=\cosh(u)\pm\sinh(u)=e^{\pm u}$
And $f(n)$ becomes $\displaystyle f(n)=\frac{e^{nu}+e^{-nu}}2=\cosh(nu)$
A similar formula arises for $a<-1$ this time $a=-\cosh(u)$ and there are $(-1)^n$ appearing.

We see coming the Tchebyshev polynomials invoked by fractal1729.
$\begin{cases}
T_n(a)=\cos(n\cos^{-1}(a)) & |a|\le 1\\
T_n(a)=\cosh(n\cosh^{-1}(a)) & a>1\\
T_n(a)=(-1)^n\cosh(n\cosh^{-1}(-a)) & a<-1\\
\end{cases}$
The case $a=1=\cos(0)$ is compatible with $\theta=0$ and $f(n)=\cos(n\times 0)=1$
The case $a=-1=\cos(\pi)$ is compatible with $\theta=\pi$ and $\cos(n\pi)=(-1)^n$
So in the end all cases except the trivial case $f=0$ can be rewritten 
$\bbox[5px,border:2px solid]{f(n)=T_n(a) \text{ where } f(1)=a}$

Edit: How to handle the case of uncertainty about $f(0)$ if we restrict the functional relation to stand only for strictly positive integers ?
The study above regarding the induction relation for $u_n$ stays valid, the only thing that changes is that we cannot determine $c$ with the initial condition $f(0)=1$ but the other condition $f(1)=a$ stands.


*

*For $a=1$ 


$f(n)=bn+c=(1-c)n+c$ 
verifying $f(m+n)+f(m-n)=(1-c)2m+2c=2f(m)\implies f(n)=1\implies c=1$


*

*For $a=-1$ 


$f(n)=(bn+c)(-1)^n=((1-c)n+c)(-1)^n$ but since $(-1)^{m+n}=(-1)^{m-n}$ 
verifying $f(m+n)+f(m-n)=2f(m)(-1)^{m+n}\implies f(n)=(-1)^n\implies c=1$


*

*For $a=0$


$f(n)=bi^n+c(-i)^n=c(i^n+(-i)^n)=2c\cos(\frac{n\pi}2)$
verifying the additive $\cos$ relation gives $4c=8c^2\implies c=\frac 12$ and $f(0)=2c=1$


*

*For $|a|>1$


$f(n)=br^n+c\bar r^n=c(r^n+\bar r^n)=2c\ T_n(a)$ and we conclude like previously that $c=\frac 12$.
Finally $f(0)$ can be determined a posteriori, and $f(0)=1$. 
A: Actually, none of your three equations must hold, because you said that the function identity only holds for positive $m$ and $n$.
So what can we say?  Well, taking $n = 1$, we see that
$$f(m+1) = 2 f(m) f(1) - f(m-1)$$
for positive $m$.  That is, the value of $f(m)$ for $m > 1$ is completely determined by the value of $f(m)$ for smaller values of $m$, which means that $f(0)$ and $f(1)$ completely determine the values of $f(m)$ for $m > 1$.
Now, taking $m = 1$, we have
$$f(1 - n) = 2 f(1) f(n) - f(n + 1)$$
for $n > 0$.  We can get any negative input we like on the right by taking a positive value of $n$, and the terms on the right will feature values of $f$ at positive inputs.  So all together this tells us that $f$ is completely determined by its values at $0$ and $1$.
Okay, so let's write $a := f(0)$ and $b := f(1)$.  We have
$$f(m+1) = 2 a f(m) - f(m-1), \qquad f(1) = b, \qquad f(0) = a$$
This is a second-order linear recurrence, just like the Fibonacci sequence.  There's a standard technique to solve these.
Once you do that you can figure out what happens for negative inputs, and then finally you can check to see which of the solutions satisfy your original functional equation in general.
