What happens to signal, if we offset periodic auto-correlation?
The periodic autocorrelation function is not defined correctly in the OP's question: it should be
$$c_n = \sum_m x_m x_{(m+n) \bmod N} ~~\text{or} ~~ \sum_m x_m x_{(m-n) \bmod N}$$ depending on taste and not $c_n = \sum_m x_mx_{(n-m)\bmod N}$ which is a convolution not a correlation operation.  Note also that there is an
implicit assumption that $\mathbf x = (x_0, x_1, \ldots, x_{N-1})$
is a real-valued sequence, not a complex-valued sequence. Of course,
the autocorrelation sequence $\mathbf c = (c_0, c_1, \ldots, c_{N-1})$ is also a real-valued
sequence with the additional property that $c_0 \geq \max_n |c_n|$

With this correction, consider the sequence 
$\mathbf d = (c_0-A, c_1-A, \ldots, c_{N-1}-A)$,
that is, $d_n = c_n-A$ for $0 \leq n &lt; N$.
Let the DFT of $\mathbf c$ be
$\mathbf C = (C_0, C_1, \ldots, C_{N-1})$ where the $C_n = |X_n|^2$ are 
nonnegative real numbers.  Note that the DFT of any
autocorrelation sequence cannot have negative values.
Now, the DFT of $\mathbf d$ is obviously
$\mathbf D = \mathbf C - (NA,0,0,\ldots, 0)$ and so we see easily that


  $\mathbf D$ cannot be the DFT of an autocorrelation function 
  if $D_0 = C_0 -NA &lt; 0$, that is, if $A &gt; N^{-1}C_0$.


Note that $\displaystyle X_0 = \sum_{n=0}^{N-1} x_n$ and
$\displaystyle C_0 = |X_0|^2 = \left|\sum_{n=0}^{N-1} x_n \right|^2.$
Let $\displaystyle \bar{x} = \frac{1}{N}\sum_{n=0}^{N-1} x_n = \frac{1}{N}X_0$.


  So, let us assume from this point on that $A \leq N^{-1}C_0 = N\left|\bar{x}\right|^2$.
  Is there
  a sequence
  $\mathbf y = (y_0, y_1, \ldots, y_{N-1})$ with
  autocorrelation function $\mathbf d$ whose DFT is $\mathbf D$?


The answer is Yes,  Consider a sequence $\mathbf y$ with
$n$-th term $y_n = x_n - a$. Its autocorrelation function
is
$$\sum_m y_m y_{(m+n)\bmod N} = \sum_m (x_m-a) (x_{(m+n)\bmod N} - a) 
= c_n + Na^2 - 2aX_0.$$
We want the quantity on the right to have value $c_n-A$, and so if
we can find an $a$ such that $A = 2aX_0- Na^2$, we are done.
Solving for $a$, we get that
$$a = \frac{2X_0\pm \sqrt{ 4\left|X_0\right|^2-4NA}}{2N}
= \frac{1}{N}X_0 \pm \sqrt{\left|\bar{x}\right|^2-A/N}
= \bar{x} \pm \sqrt{\left|\bar{x}\right|^2-A/N}$$
Note that the quantity in the radical is nonnegative since
we have assumed that $\frac AN \leq \left|\bar{x}\right|^2$
and so we do obtain a real-valued $\mathbf y$ as long as
$A$ does not exceed its maximum value $N\left|\bar{x}\right|^2$.
When $A$ does have its maximum possible
value, $\mathbf y$ whose
$n$-th term is $y_n = x_n - \bar{x}$ is just $\mathbf x$ "centered"
to have "DC value" $0$ since we are subtracting off the average
value of $\mathbf x$ from each component.