[proofplan] We prove the result by applying the Hardy-Littlewood circle method to the von Mangoldt weighted ternary representation function. The Fourier identity expresses this weighted count as an integral of a cubic exponential sum, and the major-minor arc decomposition gives a positive main term of order $N^2$ for odd $N$ while the minor arcs are $o(N^2)$. This shows that some weighted representation by prime powers exists. We then bound the total contribution from triples containing a non-prime prime power and remove it, leaving a representation by three genuine primes. [/proofplan] [step:Define the von Mangoldt weighted representation count] Let $\Lambda:\mathbb N\to\mathbb R$ denote the von Mangoldt function, so that $\Lambda(n)=\log p$ if $n=p^j$ for some prime $p$ and some $j\in\mathbb N$, and $\Lambda(n)=0$ otherwise. For each $N\in\mathbb N$, define the weighted ternary representation count $R_3:\mathbb N\to[0,\infty)$ by \begin{align*} R_3(N):=\sum_{\substack{n_1, n_2, n_3\in\mathbb N, n_1+n_2+n_3=N}}\Lambda(n_1)\Lambda(n_2)\Lambda(n_3). \end{align*} Define the additive character $e:\mathbb R\to\mathbb C$ by $e(t):=\exp(2\pi i t)$, and define the finite von Mangoldt exponential sum $S_N:\mathbb R\to\mathbb C$ by \begin{align*} S_N(\alpha):=\sum_{1\le n\le N}\Lambda(n)e(n\alpha). \end{align*} Let $\mathcal L^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb R$. By the Fourier integral identity for ternary von Mangoldt representations, [citetheorem:9079], applied with this $N$, one has \begin{align*} R_3(N)=\int_0^1 S_N(\alpha)^3 e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*} [/step] [step:Split the Fourier integral into major and minor arcs] For each sufficiently large $N$, let $\mathfrak M_N\subset[0,1)$ denote the Lebesgue measurable major-arc set supplied by the ternary Goldbach major-arc theorem [citetheorem:9082] for the sum $S_N$ and the additive character $e$, and let $\mathfrak m_N:=[0,1)\setminus\mathfrak M_N$ denote the corresponding minor-arc set used in [citetheorem:9084]. These sets are Lebesgue measurable and partition $[0,1)$, so finite additivity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} R_3(N)=\int_{\mathfrak M_N}S_N(\alpha)^3e(-N\alpha)\,d\mathcal L^1(\alpha)+\int_{\mathfrak m_N}S_N(\alpha)^3e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*} [guided] The purpose of the decomposition is to separate the frequencies where the finite von Mangoldt exponential sum $S_N:\mathbb R\to\mathbb C$ has structured arithmetic behaviour from the frequencies where cancellation dominates. Here $e:\mathbb R\to\mathbb C$ is the additive character $e(t)=\exp(2\pi i t)$, and the Fourier identity from the previous step gives \begin{align*} R_3(N)=\int_0^1 S_N(\alpha)^3e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*} For each sufficiently large $N$, the Hardy-Littlewood construction supplies a Lebesgue measurable major-arc set $\mathfrak M_N\subset[0,1)$ and its measurable complement $\mathfrak m_N=[0,1)\setminus\mathfrak M_N$. Since $\mathfrak M_N$ and $\mathfrak m_N$ are disjoint measurable subsets whose union is $[0,1)$, finite additivity of the Lebesgue integral applies to the integrable function $\alpha\mapsto S_N(\alpha)^3e(-N\alpha)$. Therefore \begin{align*} R_3(N)=\int_{\mathfrak M_N}S_N(\alpha)^3e(-N\alpha)\,d\mathcal L^1(\alpha)+\int_{\mathfrak m_N}S_N(\alpha)^3e(-N\alpha)\,d\mathcal L^1(\alpha). \end{align*} This step is not an estimate yet. It only rewrites the exact identity into the two pieces that the circle method controls separately. [/guided] [/step] [step:Use the major arc asymptotic and the minor arc bound] The sets $\mathfrak M_N$ and $\mathfrak m_N$ are chosen to be exactly the major and minor arcs in the ternary Goldbach normalization of [citetheorem:9082] and [citetheorem:9084], with the same additive character $e(t)=\exp(2\pi i t)$, the same one-dimensional Lebesgue measure $\mathcal L^1$, and the same sum $S_N(\alpha)=\sum_{1\le n\le N}\Lambda(n)e(n\alpha)$. The integer $N$ is assumed odd and sufficiently large, exactly as required by the ternary major-arc theorem. Hence the hypotheses of the major-arc theorem are satisfied. The major-arc evaluation for the ternary von Mangoldt integral, [citetheorem:9082], gives the ternary Goldbach singular series $\mathfrak S_3(N)$, defined by the Euler product/local-density formula in that theorem, and an error term $E_{\mathfrak M}(N)$ satisfying $E_{\mathfrak M}(N)=o(N^2)$ such that \begin{align*} \int_{\mathfrak M_N}S_N(\alpha)^3e(-N\alpha)\,d\mathcal L^1(\alpha)=\frac{1}{2}\mathfrak S_3(N)N^2+E_{\mathfrak M}(N). \end{align*} The same theorem includes the local-density lower bound for odd integers: there is an absolute constant $c_0>0$ such that every sufficiently large odd $N$ satisfies \begin{align*} \mathfrak S_3(N)\ge c_0. \end{align*} The minor-arc theorem [citetheorem:9084] applies to the complementary set $\mathfrak m_N$ with this same normalization; its proof uses Vinogradov's minor-arc bound for $S_N$ from [citetheorem:9083]. Therefore it gives an error term $E_{\mathfrak m}(N)$ satisfying $E_{\mathfrak m}(N)=o(N^2)$ and \begin{align*} \int_{\mathfrak m_N}S_N(\alpha)^3e(-N\alpha)\,d\mathcal L^1(\alpha)=E_{\mathfrak m}(N). \end{align*} Combining the two estimates, for odd $N$, \begin{align*} R_3(N)=\frac{1}{2}\mathfrak S_3(N)N^2+E_{\mathfrak M}(N)+E_{\mathfrak m}(N). \end{align*} Since $E_{\mathfrak M}(N)+E_{\mathfrak m}(N)=o(N^2)$ and $\mathfrak S_3(N)\ge c_0$, there exists $N_1\in\mathbb N$ such that every odd $N\ge N_1$ satisfies \begin{align*} R_3(N)\ge \frac{c_0}{4}N^2>0. \end{align*} [/step] [step:Bound the contribution from non-prime prime powers] For $N\in\mathbb N$, define the exceptional prime-power set $E_N\subset\{1,\dots,N\}$ by \begin{align*} E_N:=\{m\in\{1,\dots,N\}:m=p^j\text{ for some prime }p\text{ and some integer }j\ge 2\}. \end{align*} For each $j\ge 2$, the number of prime powers $p^j\le N$ is at most $N^{1/j}$. Hence, for $N\ge 2$, \begin{align*} \#E_N\le \sum_{2\le j\le \log_2 N}N^{1/j}\le (\log_2 N)N^{1/2}. \end{align*} Also $\Lambda(m)\le \log N$ for every $m\le N$. Therefore \begin{align*} \sum_{m\in E_N}\Lambda(m)\le N^{1/2}(\log_2 N)(\log N). \end{align*} Let $R_{3,\mathrm{bad}}(N)$ denote the part of $R_3(N)$ contributed by triples $(n_1,n_2,n_3)$ for which at least one $n_i$ belongs to $E_N$. By the union bound over the three coordinates and by the estimate $\Lambda(n)\le\log N$ for $n\le N$, \begin{align*} R_{3,\mathrm{bad}}(N)\le 3\left(\sum_{m\in E_N}\Lambda(m)\right)N(\log N)^2. \end{align*} Consequently, \begin{align*} R_{3,\mathrm{bad}}(N)\le 3N^{3/2}(\log_2 N)(\log N)^3=o(N^2). \end{align*} [/step] [step:Remove the exceptional contribution and obtain three primes] Since $R_{3,\mathrm{bad}}(N)=o(N^2)$, there exists $N_2\in\mathbb N$ such that every $N\ge N_2$ satisfies \begin{align*} R_{3,\mathrm{bad}}(N)\le \frac{c_0}{8}N^2. \end{align*} Let $N_0:=\max\{N_1,N_2\}$. If $N\ge N_0$ is odd, then \begin{align*} R_3(N)-R_{3,\mathrm{bad}}(N)\ge \frac{c_0}{8}N^2>0. \end{align*} The difference on the left is the total von Mangoldt weight of triples $(n_1,n_2,n_3)\in\mathbb N^3$ satisfying $n_1+n_2+n_3=N$ for which no $n_i$ belongs to $E_N$. Ordinary composite integers that are not prime powers may still occur in this formal sum, but their von Mangoldt weight is $0$. Since the remaining sum is positive and all its summands are non-negative, at least one remaining triple has non-zero weight; for that triple each $n_i$ has $\Lambda(n_i)>0$ and is not in $E_N$, hence each $n_i$ is a prime number. Thus there exist prime numbers $p_1,p_2,p_3\in\mathbb N$ with \begin{align*} N=p_1+p_2+p_3. \end{align*} This proves the theorem. [/step]