Question

If a function $f:Z\to Z$ is defined as $mf\left(x\right)+f\left(my\right)=f\left(f\right(x+y\left)\right)$, where $m$ is a non-zero constant, then which of the following is/are correct?

• A. $f\left(x\right)=mx+n,n\in Z$
• B. $f\left(x\right)=nx+m,n\in Z$
• C. $f\left(x\right)=0$
• D. $f\left(1\right),f\left(4\right),f\left(7\right),f\left(10\right)$ are in A.P.

Answer 1

Answer: (A), (C), (D)

The correct options are
A $f\left(x\right)=mx+n,n\in Z$
C $f\left(x\right)=0$
D $f\left(1\right),f\left(4\right),f\left(7\right),f\left(10\right)$ are in A.P.

$mf\left(x\right)+f\left(my\right)=f\left(f\right(x+y\left)\right)...\left(1\right)$
At $y=0\Rightarrow mf\left(x\right)+f\left(0\right)=f\left(f\right(x\left)\right)...\left(2\right)$
Replace $x$ by $x+1$ in eqn(2)
$mf(x+1)+f\left(0\right)=f\left(f\right(x+1\left)\right)...\left(3\right)$
Now, put $y=1$ in eqn(1), we get
$mf\left(x\right)+f\left(m\right)=f\left(f\right(x+1\left)\right)...\left(4\right)$
From eqn(3) and (4), we have
$mf(x+1)+f\left(0\right)=mf\left(x\right)+f\left(m\right)$
$\Rightarrow f(x+1)-f\left(x\right)=\frac{f\left(m\right)-f\left(0\right)}{m}=\text{Constant}$

Since, the difference of two consecutive terms $f(x+1)$ and $f\left(x\right)$ is constant.
Therefore, $f\left(1\right),f\left(4\right),f\left(7\right),f\left(10\right)$ are in A.P.

Let $f\left(x\right)=kx+n;n\in Z$
From eqn(1),
$m(kx+n)+kmy+n=f\left[k\right(x+y)+n]\Rightarrow km(x+y)+mn+n=k^2(x+y)+kn+n$

Comparing the coefficients of $(x+y)$ and constant term, we get,
$km=k^2$ and $mn+n=kn+n$
$\Rightarrow k=0,m$
When, $k=0\Rightarrow mn+n=n\Rightarrow n=0$
When, $k=m\Rightarrow mn+n=mn+n\Rightarrow n=n$
$\Rightarrow n$ can be any value.

When $k=0,n=0\Rightarrow f\left(x\right)=0,$
When $k=m,n\in Z\Rightarrow f\left(x\right)=mx+n$