= Solution
Write $e(t)=\exp(2\pi it)$. A sequence in the <circle group> is an <equidistributed sequence> if, for every interval $I\subseteq\mathbb R/\mathbb Z$,
$$
\frac1N\#\{1\leq n\leq N:\phi(n)\in I\}\longrightarrow |I|,
$$
where $|I|$ is its normalized length. Equivalently, averages of every continuous function along the sequence tend to its circle integral. <Trigonometric polynomials> approximate continuous functions, and interval indicators can be squeezed between continuous functions with arbitrarily close integrals. The nonconstant <additive characters> have integral zero. These facts give the <Weyl criterion>:
$$
\boxed{\phi\text{ is equidistributed}\iff\mathbb E_{n\leq N}e(m\phi(n))\to0\quad\text{for every }m\in\mathbb Z\setminus\{0\}.}
$$
This is the link between <equidistribution> and cancellation in <exponential sums>.
Suppose every positive-shift difference sequence were equidistributed. Fix $m\ne0$ and set $z_n=e(m\phi(n))$. For every fixed $h\geq1$, the <Weyl criterion> would give
$$
\frac1N\sum_{n=1}^{N-h}z_n\overline{z_{n+h}}\to0.
$$
The omitted final $h$ terms change a normalized average by at most $h/N$.
Here is the needed <Van der Corput inequality for finite scalar sequences>. Extend $z_n$ by zero outside $[1,N]$ and average $H$ consecutive translates of the sum. <Cauchy-Schwarz> gives
$$
\left|\frac1N\sum_{n=1}^Nz_n\right|^2
\leq\frac{N+H-1}{N}\left[\frac1H+\frac2{H^2}\sum_{h=1}^{H-1}(H-h)\operatorname{Re}\left(\frac1N\sum_{n=1}^{N-h}z_n\overline{z_{n+h}}\right)\right].
$$
Indeed, apply <Cauchy-Schwarz> to $H\sum_nz_n=\sum_t\sum_{j=0}^{H-1}z_{t-j}$ and expand the squared inner sum. Taking $N\to\infty$ first leaves a bound $1/H$; then let $H\to\infty$. Every nonzero Fourier average of $\phi$ vanishes, so the <Weyl criterion> makes $\phi$ equidistributed. This is the <differencing obstruction to equidistribution>. By contraposition, \b[a non-equidistributed sequence has a non-equidistributed difference for some positive $h$], hence for some $h\ne0$ as requested.
For $\phi(n)=\sqrt2\,n^2$, the difference is $\Delta_h\phi(n)=-2h\sqrt2\,n-h^2\sqrt2$. For any nonzero integer $m$, its <exponential sum> is a constant phase times a <geometric progression> with ratio $e(-2mh\sqrt2)\ne1$. Its normalized magnitude is at most $2/(N|1-e(-2mh\sqrt2)|)$, which tends to zero. Thus every positive-shift difference is equidistributed, and the contraposition just established proves
$$
\boxed{(\sqrt2\,n^2)_{n\geq1}\text{ is equidistributed in }\mathbb R/\mathbb Z.}
$$
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