Recent content by rrronny

Sorry... :blushing: I meant the sixth prime number. In summary, then, the only solutions (p,q) that I found are (2,2),(2,3),(3,5), (3,13).

Let v_p(\cdot) denote the p-adic valuation (see http://en.wikipedia.org/wiki/Valuation_(algebra)#p-adic_valuation"). a \nmid b \iff v_p(\frac{b}{a})=v_p(b)-v_p(a)<0 Then: v_p(\frac{b^2}{a^2})=2[v_p(b)-v_p(a)]<0 \iff a^2 \nmid b^2

\lim_{N \to \infty}P(N) = \lim_{N \to \infty} \frac{1}{N+1} = 0.

Hi CRGreathouse, thanks for your reply. Just one question: how is the trend of C(n) in the variation of n?

Hi Borek, I do not have a solution for this problem... I found only four solutions: (1,1), (1,2), (2,3), (2,6).

Let p_i denote the i-th prime number. Prove or disprove that: 1)\quad \displaystyle S(n) : = \sum_{i = 1}^n \mu(p_i + p_{i + 1}) < 0 \quad \forall n \in \mathbb{N}_0 : = \left\{1,2,3,...\right\}; 2)\quad \displaystyle S(n) \sim C \frac {n}{\log{n}}, where C is a negative real constant. In the...

Let \mathbb{P} the set of primes. Let's p,q \in \mathbb{P} and p \le q. Find the pairs (p,q) such that 2^p+3^q and 2^q+3^p are simultaneously primes.