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Work in progress:

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.

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-phase 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. In fact, coherent reception is generally not feasible in $M$-FSK systems because of the difficulty of tracking the various tones in the incoming signal in view of the fact that the tones appear and disappear from the input signal as time progresses. MSK of course is best viewed as offset QPSK with baseband pulses that are a half-period of a sinisuoid. Be that as it may, the canonical answer for $2$-FSK is $f_d = \dfrac{1}{2D}$ and so the two tones are at frequencies $f_c \pm \dfrac{1}{2D}$. The frequency spacing between the two tones in $2$-FSK is thus $\mathbf 2f_d =\dfrac{1}{D}$.

Turning to $M$-FSK with $M>2$, we have tones at $f_c + \mathbf 1f_d$ and $f_c - \mathbf 1f_d$ and so the next two tones must be at $f_c + \mathbf 3f_d$ and $f_c - \mathbf 3f_d$ in order to maintain the spacing of $2f_d$ between tones as required for orthogonality. Additional tones must be at $f_c + \mathbf nf_d$ and $f_c - \mathbf nf_d$ where $n$ is an odd number. And that's why we can't use even multiples of $f_d$ as the OP desires: the odd multiples are needed to provide the spacing of $2f_d$ between the tones.

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 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 an integer number of periods of the sinusoid, or it is the case that both RF pulses have a half-integer number of periods of the sinusoid.

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 $(f_c+f_d)D$ and $(f_c-f_d)D$ are integers, or they both are half-integers.

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.

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 $2f_d$ apart in frequency; not $f_d$ apart as the OP apparently believes. The transmitted signal consists of a succession of RF pulses of duration $D$. e 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, if the receiver consists of two matched filters for the two RF pulses and a comparator to determine which filter has 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 for details.)

Now, assuming without loss of generality that $f_1 > f_0$, we have that \begin{align} & ~~~~~~~~\int_{0}^{D}2\cos(2\pi f_1t+\theta_1)\cos(2\pi f_0t+\theta_0) \,\mathrm dt\\ &= \int_{0}^{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_{0}^{)D}\\ &= ~~~~~\frac{\sin(2\pi (f_1-f_0)D+\theta_1-\theta_0)-\sin(\theta_1-\theta_0)}{2\pi (f_1-f_0)}\\ &~~~~+\frac{\sin(2\pi (f_1+f_0)D+\theta_1+\theta_0)-\sin(\theta_1+\theta_0)}{2\pi (f_1+f_0)} \end{align} The numerators of both fractions in the last RHS above are bounded by $2$ while $2\pi (f_1+f_0)\gg 2\pi (f_1-f_0)$ and so the second fraction is considerably smaller than the first. But, if $f_1$ and $f_0$ are such that $(f_1-f_0)D$ is an integer, then the first fraction has value $0$ while if $(f_1+f_0)D$ is an integer, the second fraction has value $0$. Note that both conditions hold whenever $f_0D$ and $f_1D$ both are integers ($n$ and $m$, say) or both are half-integers ($\frac{2m-1}{2}$ and $\frac{2n-1}{2}$, say), and give us the desired orthogonality. We will restrict attention to the integer case only.

Applying the above notion to $$\int_{0}^{D}\cos(2\pi(f_c + f_d)t+\theta_1)\cos(2\pi(f_c - f_d)t+\theta_0) \,\mathrm dt = 0$$ we see that

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 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 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 an integer number of periods of the sinusoid, or it is the case that both RF pulses have a half-integer number of periods of the sinusoid.

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 $(f_c+f_d)D$ and $(f_c-f_d)D$ are integers, or they both are half-integers.