 Solution

The Vinogradov mean value isBy orthogonality of integer Fourier modes, it counts the ordered integer solutions of ${\sum }_{i=1}^k{x}_i^j={\sum }_{i=1}^k{y}_i^j$ for $1\le j\le r$, with every coordinate between one and $\lfloor Z\rfloor$. The diagonal solutions $y_i=x_i$ give $J_{k,r}(Z)\ge \lfloor Z{\rfloor }^k$.

Put $D=r(r+1)/2$. The moment vector $({\sum }_i{x}_i^j{)}_{j=1}^r$ has at most ${\prod }_{j=1}^r(k\lfloor Z{\rfloor }^j+1){\ll }_{k,r}{Z}^D$ possible values. If $R(v)$ counts the tuples with vector $v$, then ${\sum }_{v}R(v)=\lfloor Z{\rfloor }^k$ and $J_{k,r}(Z)={\sum }_{v}R(v{)}^2$. The Cauchy-Schwarz inequality gives $J_{k,r}(Z){\gg }_{k,r}{Z}^{2k-D}$. Combining the two lower bounds, with one common positive constant for large $Z$, yields

Here is an explicit way that upper bounds enter the Vinogradov mean-value method for a bilinear exponential sum. Write $U={\sum }_{x,y\le Z}e({\sum }_j{\alpha }_j{x}^j{y}^j)$ and $L_j=k\lfloor Z{\rfloor }^j$. Two applications of the Holder inequality, followed by grouping equal differences of moment vectors, giveAt integer ${\alpha }_{j}v$ the minimum is defined as $2L_j+1$; $\Vert u\Vert$ means distance to the nearest integer. To explain the mean-value factor, let $r(v)$ count pairs of $k$-tuples with prescribed moment difference. It is an autocorrelation of $R$, so $r(v)\le {\sum }_{w}R(w{)}^2=J_{k,r}(Z)$ by Cauchy-Schwarz inequality. The first Holder inequality groups the $y$ tuples, and the second groups the $x$ tuples, providing the two factors $J$. The remaining sums over moment differences are bounded by the displayed finite geometric series estimates. Good Vinogradov mean value upper bounds, together with rational approximation or spacing bounds for ${\alpha }_j$, therefore give cancellation in $U$. The mean-value estimate alone does not force cancellation for arbitrary coefficients: when all ${\alpha }_j$ are integers, $U=\lfloor Z{\rfloor }^2$. For the paper take $Z=N^{2/5}$ and the specified ${\alpha }_j$.