= Solution
The <Vinogradov mean value> is
$$
J_{k,r}(Z)=\int_{[0,1]^r}\left|\sum_{1\le z\le Z}e(\theta_1z+\cdots+\theta_rz^r)\right|^{2k}\,d\boldsymbol\theta.
$$
By <orthogonality of integer Fourier modes>, it counts the ordered integer solutions of $\sum_{i=1}^kx_i^j=\sum_{i=1}^ky_i^j$ 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 $\sum_vR(v)=\lfloor Z\rfloor^k$ and $J_{k,r}(Z)=\sum_vR(v)^2$. 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
$$
\boxed{J_{k,r}(Z)\ge c(k,r)\max\{Z^k,Z^{2k-r(r+1)/2}\}.}
$$
Here is an explicit way that upper bounds enter the <Vinogradov mean-value method for a bilinear exponential sum>. Write $U=\sum_{x,y\le Z}e(\sum_j\alpha_jx^jy^j)$ and $L_j=k\lfloor Z\rfloor^j$. Two applications of the <Holder inequality>, followed by grouping equal differences of moment vectors, give
$$
|U|^{4k^2}\ll_{k,r}Z^{8k^2-4k}J_{k,r}(Z)^2\prod_{j=1}^r\sum_{|v|\le L_j}\min\left(2L_j+1,\frac1{2\|\alpha_jv\|}\right).
$$
At integer $\alpha_jv$ the minimum is defined as $2L_j+1$; $\|u\|$ 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 $r(v)\le\sum_wR(w)^2=J_{k,r}(Z)$ 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 $\alpha_j$.
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