\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
If \(x, y\) are both non-zero, \begin {align*} \frac {1}{x} + \frac {2}{y} & = \frac {2}{7} \\ 7y + 2 \cdot 7x & = 2 xy \\ 2xy - 14x - 7y & = 0 \\ 2xy - 14x - 7y + 49 & = 49 \\ 2x (y - 7) - 7 (y - 7) & = 49 \\ (2x - 7) (y - 7) & = 49. \end {align*}
We must have \(2x - 7 \geq 2 \cdot 1 - 7 = -5\) and \(y - 7 \geq 1 - 7 = -6\).
\(2x - 7\) and \(y - 7\) are both integers, and we do casework considering expressing \(49\) into a product of two integers that are both not less than \(-6\).
Since all \(x, y\) are non-zero, we can conclude that the solutions are \((x, y) = (4, 56), (7, 14), (28, 8)\).
We have \begin {align*} p^2 + pq + q^2 & = n^2 \\ p^2 + 2pq + q^2 & = n^2 + pq \\ (p + q)^2 & = n^2 + pq \\ (p + q)^2 - n^2 & = pq \\ (p + q + n) (p + q - n) & = pq. \end {align*}
We must have \(p + q + n > p + q - n\) since \(n\) is a positive integer. We have \(p + q + n > p, q > 1 > 0\). It must be the case that \(p + q - n\) is positive as well.
Therefore, \(p + q + n\) cannot be \(1, p, q\), and it must be the case that \(p + q + n = pq\) and \(p + q - n = 1\).
Therefore, \(p + q = n + 1\), and \(pq = p + q + n = 2n + 1\).
Hence, \(p, q\) are solutions to the quadratic equation in \(t\) \[ t^2 - (n + 1)t + (2n + 1) = 0. \]
Solving this gives us \begin {align*} p, q & = \frac {(n + 1) \pm \sqrt {(n + 1)^2 - 4 \cdot (2n + 1)}}{2} \\ & = \frac {(n + 1) \pm \sqrt {n^2 - 6n - 3}}{2}. \end {align*}
We have \(n^2 - 6n - 3 = (n - 3)^2 - 12\) must be a perfect square for \(p, q\) to be rational (and they are since all integers are rational).
Consider \(a, b \geq 0\), \(a, b \in \NN \) such that \(a^2 - b^2 = (a + b) (a - b) = 12\).
\(a + b\) and \(a - b\) must take the same odd-even parity, and the only possibility is therefore \(a + b = 6\) and \(a - b = 2\), solving to \((a, b) = (4, 2)\).
Therefore, \(n - 3 = 4\), \(n = 7\), and we solve for \[ p, q = \frac {8 \pm \sqrt {49 - 42 - 3}}{2} = 4 \pm 1 \] and \((p, q) = (3, 5), (5, 3)\) are indeed primes, and \(n = 7\).
If \(p + q - n \geq p\), then \(q \geq n\), and for the original equation, \[ \LHS = p^3 + q^3 + 3pq^2 > q^3 \geq n^3 = \RHS , \] and hence \(\LHS > \RHS \) is impossible. Hence, \(p + q - n < p\).
It must also be the case for \(p + q - n < q\).
We have \begin {align*} p^3 + q^3 + 3pq^2 & = n^3 \\ p^3 + q^3 + 3pq^2 + 3p^2 q & = n^3 + 3p^2 q \\ (p + q)^3 & = n^3 + 3p^2 q \\ (p + q)^3 - n^3 & = 3p^2 q \\ (p + q - n) \left [(p + q)^2 + (p + q) \cdot n + n^2\right ] & = 3p^2 q. \end {align*}
The factors of \(3p^2 q\) are (given \(p\) and \(q\) are prime), \[ 1, 3, p, q, 3p, 3q, p^2, pq, 3p^2, 3pq, p^2 q, 3 p^2 q, \] and since \(p + q - n < p\) and \(p + q - n < q\), it must be either the case that \(p + q - n = 1\) or \(p + q - n = 3\).
If \(p + q - n = 1\), then \(p + q = n + 1\), we have \begin {align*} (p + q)^2 + (p + q) n + n^2 & = 3 p^2 q \\ (n + 1)^2 + (n + 1) n + n^2 & = 3 p^2 q \\ 3n^2 + 3n + 2 & = 3 p^2 q. \end {align*}
The left-hand side is congruent to \(1\) modulo \(3\), while the right-hand side is a multiple of \(3\), so this is impossible.
If \(p + q - n = 3\), \(p + q = n + 3\), we have \begin {align*} (p + q)^2 + (p + q) n + n^2 & = p^2 q \\ (n + 3)^2 + (n + 3)n + n^2 & = p^2 q \\ 3n^2 + 9n + 9 & = p^2 q \\ 3 (n^2 + 3n + 3) & = p^2 q. \end {align*}
Therefore, \(3 \divides p^2 q\), and hence \(3 \divides p\) or \(3 \divides q\), and hence either \(p\) or \(q\) must be \(3\) and the other one is \(n\). However, we have concluded that \(p + q - n < p \iff q < n\) and \(p + q - n < q \iff p < n\), which makes this impossible.
This shows that it is impossible for primes \(p, q\) and integer \(n\) such that \(p^3 + q^3 + 3pq^2 = n^3\), which shows that there are no primes \(p, q\) such that \(p^3 + q^3 + 3pq^2\) is the cube of an integer.