From + To -

Let $P(m,n)$ be the statement $f(m+f(n)) = f(m)-n$ .

$P(0,n) \implies f(f(n)) = f(0)-n$ . If for some $a,b$ we have $f(a) = f(b)$ , then $a = f(0) - f(f(a)) = f(0) - f(f(b)) = b$ , so $f$ is injective.

$P(m,0) \implies f(m+f(0)) = f(m)$ . Since $f$ is injective, $m+f(0) = m$ , so $f(0) = 0$ .

$P(-f(n), f(n)) \implies f(0) = f(-f(n)) - n$ or $f(-f(n)) = n$ .

$P(m,-f(n)) \implies f(m+f(-f(n))) = f(m) + f(n)$ , or $f(m+n) = f(m) + f(n)$ . This is the well known Cauchy functional equation, whose solution is $f(x) = ax$ for some integer $a$ . However, plugging this back, we find that this doesn't satisfy the equation for any $a$ : $f(m+f(n)) = a(m+a(n)) = am + a^2 n$ , while $f(m)-n = am - n$ , which means we need $a^2 = -1$ . So there is no solution.

I thought in the following way.Not sure it is right or wrong.

$f(m+f(n))=f(m)-n$

Putting $-n$ in place of $n$ equation becomes:

$f(m+f(-n))=f(m)+n$

Subtracting two equations :

$f(m+f(-n))-f(m+f(n))=2 n$ .

Putting $m-0$

$f(f(-n))-f(f(n))=2 n$

It is clear that $f(n)$ is a polynomial function.Let $f(n)$ has degree $l$ .Then $f(f(n))$ has degree $2 l$ The above equation gives a residue of degree 1 polynomial.The even will cancels out in subtraction.So $f(f(n))$ can have maximum powers of 2 then $f(n)$ can have degree $1$

After checking it may be seen that there exists no function.