Let $p$ and $q$ be integers greater than $1$, and let $pq$ denote their product. Recall that the quantum Fourier transform modulo $pq$ is the $pq$-dimensional unitary operation defined as
\[
F_{pq}\ket{x}=\frac{1}{\sqrt{pq}}\SUM{y=0}{pq-1}\omega_{pq}^{xy}\ket{y},
\]
for each $x\in\mathbb{Z}_{pq}$, where $\omega_{N}=e^{i2\pi/N}$.
Define two quantum states $\ket{\psi_1}$ and $\ket{\psi_2}$ as
\[
\ket{\psi_1}=\frac{1}{\sqrt{q}}\SUM{x=0}{q-1}\ket{xp} \ \ \ \text{and} \ \ \ \ket{\psi_2}=\frac{1}{\sqrt{p}}\SUM{x=0}{p-1}\ket{xq}.
\]
Then
\[\begin{align*}
F_{pq}\ket{\psi_1}&=\frac{1}{q\sqrt{p}}\SUM{x=0}{q-1}\SUM{y=0}{pq-1}\omega_{pq}^{xpy}\ket{y} \\
&=\frac{1}{q\sqrt{p}}\SUM{y=0}{pq-1}\left(\SUM{x=0}{q-1}\omega_{q}^{xy})\right)\ket{y} \\
&=\frac{1}{q\sqrt{p}}\SUM{n=0}{p-1}\left(\SUM{x=0}{q-1}1\right)\ket{nq} \\
&=\frac{1}{q\sqrt{p}}\SUM{n=0}{p-1}q\ket{nq} \\
&=\frac{1}{\sqrt{p}}\SUM{n=0}{p-1}\ket{nq} \\
&=\ket{\psi_1},
\end{align*}\]
where the third and fourth lines follows from the general fact that
\[\SUM{x=0}{N-1}\omega_{N}^{xy}=\left\{
\begin{array}{ll} N \ &\text{if} \ y=0 (\text{mod} \ N) \\
0 &\text{if} \ y\neq 0 (\text{mod} \ N)
\end{array}
\right.,
\]
and that $y=0 (\text{mod} \ q)$ implies that $y=nq$ for some integer $n$.
Let $s=\{0,1,\dots, p-1\}$, and define $\ket{\psi_3}$ (a ``shifted" version of $\ket{\psi_1}$) as
\[\ket{\psi_3}=\frac{1}{\sqrt{q}}\SUM{x=0}{q-1}\ket{s+xp}.\]
Then
\[\begin{align*}
F_{pq}\ket{\psi_3}&=\frac{1}{q\sqrt{p}}\SUM{x=0}{q-1}\SUM{y=0}{pq-1}\omega_{pq}^{(s+xp)y}\ket{y} \\
&=\frac{1}{q\sqrt{p}}\SUM{x=0}{q-1}\SUM{y=0}{pq-1}\omega_{pq}^{sy}\omega_{pq}^{xpy}\ket{y} \\
&=\frac{1}{q\sqrt{p}}\SUM{y=0}{pq-1}\omega_{pq}^{sy}\SUM{x=0}{q-1}\omega_{q}^{xy}\ket{y} \\
&=\frac{1}{q\sqrt{p}}\SUM{n=0}{p-1}\omega_{pq}^{snq}\SUM{x=0}{q-1}1\ket{nq} \\
&=\frac{1}{q\sqrt{p}}\SUM{n=0}{p-1}\omega_{p}^{sn}q\ket{nq} \\
&=\frac{1}{\sqrt{p}}\SUM{n=0}{p-1}\omega_{pq}^{sn}\ket{nq}.
\end{align*}\]
Let $\alpha_{y}$ denote the amplitude of the state $\ket{y}$ in the super position described in $F_{pq}\ket{\psi_1}$ just derived. Since any state $\ket{y}\neq\ket{nq}$ for some $n\in\{0,1,\dots,p-1\}$ does not appear in the superposition $F_{pq}\ket{\psi_1}$, this implies that the amplitude $\alpha_y$ for such states is $0$ and will be observed with probability $0$. On the contrary, for any state $\ket{y}\neq\ket{nq}$ for some $n\in\{0,1,\dots,p-1\}$, the corresponding amplitude is given by $\alpha_{y}=\frac{1}{\sqrt{p}}\omega_{pq}^{sn}$. Therefore the probability of observing the state $\ket{y}\neq\ket{nq}$ is $|\alpha_y|^2=\frac{1}{\sqrt{p}}\omega_{pq}^{sn}\frac{1}{\sqrt{p}}\overline{\omega}_{pq}^{sn}=\frac{1}{p}$. Hence, if $F_{pq}\ket{\psi_1}$ is measured in the computation basis, only some state $\ket{nq}$ representing an integer multiple of $q$ will be observed with uniform probability of $\frac{1}{p}$.
\[
F_{pq}\ket{x}=\frac{1}{\sqrt{pq}}\SUM{y=0}{pq-1}\omega_{pq}^{xy}\ket{y},
\]
for each $x\in\mathbb{Z}_{pq}$, where $\omega_{N}=e^{i2\pi/N}$.
Define two quantum states $\ket{\psi_1}$ and $\ket{\psi_2}$ as
\[
\ket{\psi_1}=\frac{1}{\sqrt{q}}\SUM{x=0}{q-1}\ket{xp} \ \ \ \text{and} \ \ \ \ket{\psi_2}=\frac{1}{\sqrt{p}}\SUM{x=0}{p-1}\ket{xq}.
\]
Then
\[\begin{align*}
F_{pq}\ket{\psi_1}&=\frac{1}{q\sqrt{p}}\SUM{x=0}{q-1}\SUM{y=0}{pq-1}\omega_{pq}^{xpy}\ket{y} \\
&=\frac{1}{q\sqrt{p}}\SUM{y=0}{pq-1}\left(\SUM{x=0}{q-1}\omega_{q}^{xy})\right)\ket{y} \\
&=\frac{1}{q\sqrt{p}}\SUM{n=0}{p-1}\left(\SUM{x=0}{q-1}1\right)\ket{nq} \\
&=\frac{1}{q\sqrt{p}}\SUM{n=0}{p-1}q\ket{nq} \\
&=\frac{1}{\sqrt{p}}\SUM{n=0}{p-1}\ket{nq} \\
&=\ket{\psi_1},
\end{align*}\]
where the third and fourth lines follows from the general fact that
\[\SUM{x=0}{N-1}\omega_{N}^{xy}=\left\{
\begin{array}{ll} N \ &\text{if} \ y=0 (\text{mod} \ N) \\
0 &\text{if} \ y\neq 0 (\text{mod} \ N)
\end{array}
\right.,
\]
and that $y=0 (\text{mod} \ q)$ implies that $y=nq$ for some integer $n$.
Let $s=\{0,1,\dots, p-1\}$, and define $\ket{\psi_3}$ (a ``shifted" version of $\ket{\psi_1}$) as
\[\ket{\psi_3}=\frac{1}{\sqrt{q}}\SUM{x=0}{q-1}\ket{s+xp}.\]
Then
\[\begin{align*}
F_{pq}\ket{\psi_3}&=\frac{1}{q\sqrt{p}}\SUM{x=0}{q-1}\SUM{y=0}{pq-1}\omega_{pq}^{(s+xp)y}\ket{y} \\
&=\frac{1}{q\sqrt{p}}\SUM{x=0}{q-1}\SUM{y=0}{pq-1}\omega_{pq}^{sy}\omega_{pq}^{xpy}\ket{y} \\
&=\frac{1}{q\sqrt{p}}\SUM{y=0}{pq-1}\omega_{pq}^{sy}\SUM{x=0}{q-1}\omega_{q}^{xy}\ket{y} \\
&=\frac{1}{q\sqrt{p}}\SUM{n=0}{p-1}\omega_{pq}^{snq}\SUM{x=0}{q-1}1\ket{nq} \\
&=\frac{1}{q\sqrt{p}}\SUM{n=0}{p-1}\omega_{p}^{sn}q\ket{nq} \\
&=\frac{1}{\sqrt{p}}\SUM{n=0}{p-1}\omega_{pq}^{sn}\ket{nq}.
\end{align*}\]
Let $\alpha_{y}$ denote the amplitude of the state $\ket{y}$ in the super position described in $F_{pq}\ket{\psi_1}$ just derived. Since any state $\ket{y}\neq\ket{nq}$ for some $n\in\{0,1,\dots,p-1\}$ does not appear in the superposition $F_{pq}\ket{\psi_1}$, this implies that the amplitude $\alpha_y$ for such states is $0$ and will be observed with probability $0$. On the contrary, for any state $\ket{y}\neq\ket{nq}$ for some $n\in\{0,1,\dots,p-1\}$, the corresponding amplitude is given by $\alpha_{y}=\frac{1}{\sqrt{p}}\omega_{pq}^{sn}$. Therefore the probability of observing the state $\ket{y}\neq\ket{nq}$ is $|\alpha_y|^2=\frac{1}{\sqrt{p}}\omega_{pq}^{sn}\frac{1}{\sqrt{p}}\overline{\omega}_{pq}^{sn}=\frac{1}{p}$. Hence, if $F_{pq}\ket{\psi_1}$ is measured in the computation basis, only some state $\ket{nq}$ representing an integer multiple of $q$ will be observed with uniform probability of $\frac{1}{p}$.