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Use the integer norm lower bound for rational approximation to a quadratic irrational. The integer $2q^2-a^2$ is nonzero, since $\sqrt{2}$ is irrational. If $a/q<2$, thenbecause $\sqrt{2}+a/q<\sqrt{2}+2<4$. If $a/q\ge 2$, the difference is at least $2-\sqrt{2}>1/4\ge 1/(4q^2)$. ThusIn particular, $\sqrt{2}$ 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 $a/q$ withCombining 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:Here is the structure behind that estimate. With Dirichlet convolution and cutoffs $U,V$, the Vaughan identity isIt follows from $\Lambda =\mu \ast \log$, $\log =\Lambda \ast 1$, and $\mu \ast 1={\delta }_1$. 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 $\min (L,(2\Vert \alpha m\Vert {)}^{-1})$. 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 $\alpha =\sqrt{2}$ and $q\asymp N^{1/2}$, the two denominator-dependent terms are $O(N^{3/4})$. ConsequentlyThe final passage absorbs the logarithm using the positive exponent margin $9/10-4/5=1/10$.