Half of primes have the even sum of digits

You need to know: Prime numbers, relatively prime integers.

Background: Let $q\geq 2$ be an integer. Any integer $n>0$ can be uniquely represented as $n=\sum\limits_{i=1}^{k} b_i q^{k-i}$ , where $k>0$ is an integer and $b_1, b_2, \dots, b_k$ are integers between $0$ and $q-1$ , which are called digits of n in the q-ary number system (if $q=10$ , these are the digits in the usual decimal system). Let $S_q(n)=\sum\limits_{i=1}^k b_i$ be the sum of digits of n. Let $\pi(n)$ be the number of primes less than n, and let $\pi_{q,m,a}(n)$ be the number of primes p less than n such that $S_q(p) - a$ is a multiple of m.

The Theorem: On 10th November 2005, Christian Mauduit and Joël Rivat submitted to the Annals of Mathematics a paper in which they proved that for all integers $m \geq 2$ and $q \geq 2$ such that $q-1$ and $m$ are relatively prime, there exist constants $C_{q,m}>0$ and $\sigma_{q,m}>0$ , such that the inequality $\left|\pi_{q,m,a}(n) - \frac{\pi(n)}{m}\right| \leq C_{q,m} n^{1-\sigma_{q,m}}$ holds for all integers $n>0$ and all integers $a$ .

Short context: Because $C_{q,m} n^{1-\sigma_{q,m}}$ is known to be negligible comparing to $\pi(n)$ for large n, the theorem states that $\pi_{q,m,a}(n) \approx\frac{\pi(n)}{m}$ , that is, the primes whose sum of digits gives any fixed reminder after division by m occupies about $1/m$ of all primes, as intuitively expected. This answers the 1968 question of Gelfond. In the special case $q=10$ and $m=2$ , it states that asymptotically half of all primes have the even sum of digits, and half of primes have the odd sum of digits.