Beautiful FE

[asif e elahi]

Let $n$ be an positive integer.Find all functions $f$ from integers to integers and $f(x+y+f(y))=f(x)+ny$ for all integers $x$ and $y$.

[Fm Jakaria]

The given equation is denoted $(M)$.
First we note that if $f(a) = f(b)$, $(M)$ implies
$nb = f(a+b+f(b)) – f(a) = f(b+a+f(a))- f(b) = na \Rightarrow a = b$, as $n \neq 0$.
$\rightarrow f$ is injective.
Now $(M) y = 0 \rightarrow f(x+f(0)) = f(x) \rightarrow f(0) = 0$.
($M) x=0 \rightarrow f(y+f(y)) = ny ….(1)$
$(M) x = -y \rightarrow f(f(y)) – ny = f(-y)….(2)$
$(M) x = -f(y) \rightarrow f(y) = f(-f(y)) + ny…(3)$
$(3)(-y) \rightarrow f(-y) = f(-f(-y)) – ny…(3_)$
$(3_),(2) \rightarrow f(f(y)) = f(-f(-y)) \rightarrow f(-y) = -f(y)….(ii)$
$(2),(ii) \rightarrow f(f(y)) + f(y) = ny….(2_)$
$(2_) y+f(y) \rightarrow (1) \rightarrow f(ny) + ny = ny + nf(y)$
$\Rightarrow f(ny) = nf(y)….(iii)$

$(M) nx,y \rightarrow (iii) \rightarrow f(nx+y+f(y)) = n(f(x)+y)$
Set $y = f(z)-f(x)$ for any integer $z$ in the last equation to get,
$\rightarrow f(nx + f(z) – f(x) + f(f(z)-f(x))) = nf(z) \rightarrow (iii) f(nz)$
$\Rightarrow f(z) – f(x) + f(f(z)-f(x)) = n(z-x)$
$\Rightarrow f(f(z) – f(x) + f(f(z)-f(x))) = f(n(z-x))$
Using $(1) y= f(z)-f(x)$, LHS is $n(f(z) – f(x))$. Using (iii), RHS is $nf(z-x)$.
As $n \neq 0$, $f(z) - f(x) = f(z-x)$.
Now in the last relation we input $–x$ instead of $x$ to get, using $(ii)$;
$f(x) + f(z) = f(x + z)$; for all integers $x,z$.
As $f$’s domain and range is included in integers, we conclude that if $f(1) = m$ is a fixed integer; then $f(x) = mx..(C)$ for all integer $x$.
From this, we choose any integer $x$ and any nonzero integer $y$ in $(M)$ to get;
$n = m(m+1)…(I)$
Conversely, if $(I)$ holds for some fixed integer $m$; this function solves $(M)$.

So our answer is; if $(I)$ do not have a integer solution for some positive integer $n$; there’s no solution.
Conversely, if $(I)$ has a solution; solve it in integers $m$ to get two distinct solutions $f $ by $(C)$, corresponding to:
$m = \frac{-1 \pm \sqrt {1+4n}}{2}$. [This happens precisely when $4n+1$ is a perfect square]