= Solution
Use the <integer norm lower bound for rational approximation to a quadratic irrational>. The <integer> $2q^2-a^2$ is nonzero, since $\sqrt2$ is irrational. If $a/q<2$, then
$$
\left|\sqrt2-\frac aq\right|
=\frac{|2q^2-a^2|}{q^2(\sqrt2+a/q)}
>\frac1{4q^2},
$$
because $\sqrt2+a/q<\sqrt2+2<4$. If $a/q\geq2$, the difference is at least $2-\sqrt2>1/4\geq1/(4q^2)$. Thus
$$
\boxed{\left|\sqrt2-\frac aq\right|>\frac1{4q^2}\quad(a,q\geq1).}
$$
In particular, $\sqrt2$ is a <badly approximable number>.
To estimate the prime-weighted <exponential sum>, let $Q=\lfloor N^{1/2}\rfloor$. The <Dirichlet approximation theorem> gives a reduced fraction $a/q$ with
$$
1\leq q\leq Q,\qquad
\left|\sqrt2-\frac aq\right|\leq\frac1{qQ}\leq\frac1{q^2}.
$$
Combining this with the lower bound gives $q>Q/4$. Thus $q\asymp N^{1/2}$; the approximation cannot have a small denominator.
The relevant analytic input is the <prime exponential-sum estimate from Vaughan identity>:
$$
\sum_{n\leq N}\Lambda(n)e(\alpha n)
\ll\left(\frac N{\sqrt q}+N^{4/5}+\sqrt{Nq}\right)\log^4(2N)
\quad\left(\left|\alpha-\frac aq\right|\leq q^{-2},\ (a,q)=1\right).
$$
Here is the structure behind that estimate. With <Dirichlet convolution> and cutoffs $U,V$, the <Vaughan identity> is
$$
\Lambda=\mu_{\leq U}*\log
-\mu_{\leq U}*\Lambda_{\leq V}*1
+\Lambda_{\leq V}
+\mu_{>U}*\Lambda_{>V}*1.
$$
It follows from $\Lambda=\mu*\log$, $\log=\Lambda*1$, and $\mu*1=\delta_1$. The first two <Dirichlet convolution> terms become <type I sums> with one short factor; the last becomes a bilinear <type II sum> with both factors bounded below. Split these sums into dyadic blocks and use $U=V=N^{2/5}$. For the <type I sums>, the inner <finite geometric series> has bound $\min(L,(2\|\alpha m\|)^{-1})$. For the <type II sums>, the <Cauchy-Schwarz inequality> followed by expansion introduces analogous sums in differences of indices. Rational approximation separates the phases into blocks of length comparable to $q$, bounding these reciprocal-distance sums. The <Dirichlet convolution> coefficients are divisor-bounded, producing the logarithmic factors. These steps give the displayed estimate; no unproved cancellation of arbitrary weighted sums is being assumed.
With $\alpha=\sqrt2$ and $q\asymp N^{1/2}$, the two denominator-dependent terms are $O(N^{3/4})$. Consequently
$$
\boxed{\sum_{n\leq N}\Lambda(n)e(n\sqrt2)
=O(N^{4/5}\log^4(2N))=O(N^{9/10}).}
$$
The final passage absorbs the logarithm using the positive exponent margin $9/10-4/5=1/10$.
Back to article page
