Primes in the sequences $1+3n$ and $1+4n$ I'm studying primes in two sequences. By analogy with the Chebychev's work, define the functions
$$\psi_*(x)= \sum_{n\leq x} \Lambda(1+3n)\Lambda(1+4n)$$
and 
$$\theta_*(x) = \sum_{n\leq x}\log(1+3n)\log(1+4n).$$
How to prove that, for some constant $C$, 
$$\psi_*(x)\sim Cx\iff \theta_*(x)\sim Cx.\;\;\;\;\;\;\;\;\;\;\; (I)$$ 
I tried to estimate the difference $\psi_*(x)-\theta_*(x)$, but I got stuck on the counting of the number of $n\leq x$ such that $1+4n$ and $1+3n$ are both power of primes, say $p^k$ and $q^l$, with $k,l\geq 1$ and $kl>1$. 
I also tried to prove that $\psi_*(x) = O(x)$, without success. Can someone help prove (I) and $\psi_*(x) = O(x)$.
 A: The upper bound of Proposition 5.6 is the heart of this proof. 
Theorem 5.7. $\psi_f(x)\sim C(f)x \iff \theta_f(x)\sim C(f)x. $
The polynomials $f_1= 3n+1,f_2=4n+1 $ satisfy the requirements of 2.2 of the paper. $\psi$ runs over all powers of primes while $\theta$ runs only over primes. The difference is a sum 
$$\psi_f(x)-\theta_f(x) = \sum_{n\leq x}\Lambda(f_1(n))\Lambda_2(f(n)) $$
over products $\log(3n+1)\log(4n+1)$ such that both polynomials represent a prime power $p^k, k>1.$ 
The second half of the theorem incorporates an earlier argument about size (5.6). The first step is the inequality:
$$\sum\Lambda(3n+1)\Lambda(4n+1)\leq \log(3x+1)\log(4x+1)\sum_{n\leq x} 1$$
where again the sum on both sides is over prime powers greater than 1. For the next version of the inequality we have to return to a definition on page 40. 

Definition. Let $f_0(X)\in Z[X]$ and $k\in N,k\geq 2.$ We define 
$$Z_k(f_0;x)~\dot{=}~~\#\{n\in N, n\leq x:\exists ~m\in Z ~s.t.~ f_0 = m^k \} $$  
This gives the cardinality of values for which a polynomial $f$ equals a prime power with exponent $k.$
We can re-write the sum:
$$(1) \hspace{10mm}\psi_f(x)-\theta_f(x)\leq \log(f_1(x))\cdot\log (f_2(x))\sum_{i=1}^2\sum_{m=2}^{\log_2f_i(x)}~Z_m(f_i;x) $$  
The theorem says that using (1) and 5.6 we can deduce that 
$$(2)\hspace{7mm}\psi_f(x)-\theta_f(x) = O(\sqrt{ x}\log^2 x).  $$
Using 5.6 we can bound the number of prime powers instead of counting them. This turns out to be enough.

Here is a rough translation of 5.6 leaving out cases for higher degree polynomials. 
Proposition 5.6. Let $f_0(X) = a_dX^d+...+a_0\in Z[X]$ be irreducible, strictly increasing on $(-1,\infty)$ and with $f_0(x)>0.$ Then there exists an $M>0$ (depending on $f_0(X)$) such that given $k\in N, k\geq 2,$
$$Z_k(f_0;x)\leq M\sqrt{x}. $$ 
Proof. $d$ is the degree of the polynomial, in this case $d =1.$ In this case we always have $k$ (from definition of $Z_k$)$~~\geq 2d$ and the result follows as per (a) of case 1. 
Case (1)(a).   $k\geq 2d.$ For $x\geq 1, n \in N, n \leq x,$ we have 
$$\sqrt[k]{f_0(n)}\leq \sqrt[k]{|a_1|n+|a_0|}\leq \sqrt[k]{|a_1|} n^{1/k}+\sqrt[k]{|a_0|} $$
$$\leq |a_1|n^{1/k}+|a_0|\leq (|a_1|+|a_0|)n^{1/k}\leq M_1\sqrt{x} $$
in which we use $1/k \leq 2,^*$ and $M_1:= |a_1|+|a_0|$ depends on $f_0(x).$ In this way
if $\sqrt[k]{f_0(n)}$ is an integer it will lie between 1 and $[M_1\sqrt{x}],$ the integer part of $M_1\sqrt{x},$ so there will be $[M_1\sqrt{x}]$ possibilities for values of $\sqrt[k]{f_0(n)}.$ Since $n\rightarrow f_0(n)$ is injective by hypothesis, we can have at maximum $[M_1\sqrt{x}]$ values of n 
 for which $\sqrt[k]{f_0(n)}$ is a whole number. Therefore $Z_k(f_0;x)\leq M_1\sqrt{x}.$
Notice that a sum of such bounds will also have the form $M\sqrt{x}$ and the source of the factor $\log^2 x$ in (2) should be clear enough.

If $\psi(x)-\theta(x) = O(\sqrt{x}\log^2 x)$ and we want to show that$^{**}$ 
$$\psi(x)\sim Cx\iff \theta(x)\sim Cx, $$ 
assume first that $\psi(x)\sim Cx. $ Then by definition of $O(x)$
$$\left|\frac{\psi(x)-\theta(x)}{\sqrt{x}\log^2 x}\right|\leq M $$ for some positive constant M. Both numerator and denominator are positive. We can multiply both by $1/Cx.$
$$\frac{\psi(x)}{Cx}-\frac{\theta(x)}{Cx}\leq M \left(\frac{\sqrt{x}\log^2 x}{Cx}\right)$$
and as $x\to \infty $ the equivalence $\theta(x)\sim Cx$ follows (and conversely). 
$^{*}$I think this should be $1/k\leq 1/2.$
$^{**}$ Using $\psi$ here for $\psi_f$. 
