**Work in progress:**


>So, my question is: why can't do we use any value (even and odd, spaced by 1 fd), to reduce even more the bandwidth of our signal? Has it relation to interference between symbols in frequency spectrum?

Let's first consider _binary_ or $2$-FSK. We have a carrier signal at fixed frequency $f_c$. _Modulation_ by the binary data changes the frequency of the transmitted signal to $f_c + f_d$ **or** to $f_c-f_d$. Thus the _change_ in the frequency of the transmitted signal which occurs when the transmitter transmits a $1$ followed by a $0$ is $\pm 2f_d$; the two possible FSK tones at frequencies $f_c + f_d$ and $f_c-f_d$ are spaced $\mathbf 2f_d$ apart in frequency; _not_ $f_d$ apart as the OP apparently believes. (The "$f_d$ apart" model corresponds ta a different system where to send a $0$, a tone at $f_c$ is transmitted whereas to send a $1$, a tone at $f_c+f_d$ is transmitted). The transmitted signal consists of a succession of RF pulses of duration $D$. Let's  assume that $f_c \gg \dfrac 1D$ so that each RF pulse has several periods of the RF sinusoid. 

What we would like to arrange is for the two possible tones to be _orthogonal_ over each signaling epoch, that is, the RF pulses
$\cos(2\pi(f_c + f_d)t+\theta_1)$ and $\cos(2\pi(f_c - f_d)t+\theta_0)$ should satisfy
$$\int_{kD}^{(k+1)D}\cos(2\pi(f_c + f_d)t+\theta_1)\cos(2\pi(f_c - f_d)t+\theta_0) \,\mathrm dt = 0$$
for all integers $k$.  Why the insistence on orthogonality? Well, suppose that the receiver consists of two _matched filters_ for the two RF pulses and a comparator to determine which filter is producing a larger response at the sampling instant. Then, the output of each matched filter is the cross-correlation function of the input RF pulse with the RF pulse to which the filter is matched, and at the sampling instant, this cross-correlation has value $0$ if the input is the _other_ pulse, and large value (the peak value of the autocorrelation function, in fact) if the input is the one to which the filter is matched (cf. [this answer](https://dsp.stackexchange.com/a/9389/235) for details.)

Now, assuming without loss of generality that $f_1 > f_0$, we have that
\begin{align} 
& ~~~~~~~~\int_{kD}^{(k+1)D}2\cos(2\pi f_1t+\theta_1)\cos(2\pi f_0t+\theta_0) \,\mathrm dt\\
&= \int_{kD}^{(k+1)D}\cos(2\pi (f_1-f_0)t+\theta_1-\theta_0) +\cos(2\pi (f_1+f_0)t+\theta_1+\theta_0) \,\mathrm dt\\
&= \left.\frac{\sin(2\pi (f_1-f_0)t+\theta_1-\theta_0)}{2\pi (f_1-f_0)}
+  \frac{\sin(2\pi (f_1+f_0)t+\theta_1+\theta_0)}{2\pi (f_1+f_0)}\right\vert_{kD}^{(k+1)D}\\
&= ~~~~~\frac{\sin(2\pi (f_1-f_0)(k+1)D+\theta_1-\theta_0)-\sin(2\pi (f_1-f_0)kD+\theta_1-\theta_0)}{2\pi (f_1-f_0)}\\
&~~~~+\frac{\sin(2\pi (f_1+f_0)(k+1)D+\theta_1+\theta_0)-\sin(2\pi (f_1+f_0)kD+\theta_1+\theta_0)}{2\pi (f_1+f_0)}
\end{align}
The numerators of the two fractions in the last RHS above both have values in $[-2,+2]$ while in the denominators we have that $2\pi (f_1+f_0)\gg 2\pi (f_1-f_0)$. So the magnitude of the second fraction is small in comparison to the magnitude of the first. Now, bearing in mind that $k$ and $k+1$ are integers, let us note that if $f_1$ and $f_0$ are such that $(f_1-f_0)D$ is an integer, then the first fraction has value $0$. Similarly, if $f_1$ and $f_0$ are such that $(f_1+f_0)D$ is an integer, then the second fraction has value $0$. Now, if $f_1D$ and $f_0D$ both are distinct) integers ($m$ and $n$, say, with $m> n$) or both are _half-integers_ ($\frac{2m-1}{2}$ and $\frac{2n-1}{2}$, say), then both $(f_1-f_0)D$ and $(f_1+f_0)D$ are integers, and the two possible RF pulses at frequencies $f_1$ and $f_0$ are orthogonal over the interval $[kD, (k+1)D]$. Note that in all symbol epochs $[kD,(k+1)D]$, it is either the case that both RF pulses have (different) integer number of periods of their respective sinusoids, or it is the case that both RF pulses have different  half-integer number of periods of their respective sinusoids.

Applying the above notion to
$$\int_{kD}^{(k+1)D}\cos(2\pi(f_c + f_d)t+\theta_1)\cos(2\pi(f_c - f_d)t+\theta_0) \,\mathrm dt = 0$$
we see that it must be that both $(f_1-f_0)D = 2f_dD$ and $(f_1+f_0)D = 2f_cD$ are distinct integers, or they both are distinct half-integers. The smallest possible value of $2f_D$ is thus $\frac 12$ which gives $f_d = \dfrac{1}{4D}$, which will readily be recognized as the frequency deviation in _minimum-shift keying_ (MSK), which, though technically a special form of _continuous-phas_ FSK or CPFSK, is _not_ what is generally meant in general discussions of $2$-FSK or $M$-FSK which are often geared towards systems where _noncoherent_ receivers are often used.