\(% 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}}\)
In the \(n + k\) integers, the first one is \(c\), and the final one is \(c + n + k - 1\).
In the \(n\) integers, the first one is \(c + n + k\), and the final one is \(c + 2n + k - 1\).
Hence, the sums are equal if and only if \begin {align*} \frac {(n + k) [c + (c + n + k - 1)]}{2} & = \frac {n [(c + n + k) + (c + 2n + k - 1)]}{2} \\ (n + k) (2c + n + k - 1) & = n (2c + 3n + 2k - 1) \\ 2cn + n^2 + nk - n + 2ck + kn + k^2 - k & = 2cn + 3n^2 + 2kn - 1 \\ 2ck + k^2 & = 2n^2 + k, \end {align*}
as desired. All the above steps are reversible.
When \(k = 1\), \(2c + 1 = 2n^2 + 1\), and \(c = n^2\).
Hence, \[ (c, n) \in \left \{(t^2, t) \mid t \in \NN \right \}, \] and \(n\) can take all positive integers.
When \(k = 2\), \(4c + 4 = 2n^2 + 2\), and \(2c = n^2 - 1\).
By parity, \(n\) must be odd. Let \(n = 2t - 1\) for \(t \in \NN \), and we have \[ 2c = (2t - 1)^2 - 1 = 4t^2 - 4t, \] and hence \[ c = 2t^2 - 2t. \]
Hence, \[ (c, n) \in \left \{(2t^2 - 2t, 2t - 1) \mid t \in \NN \right \}, \] and \(n\) can take all odd positive integers.
If \(k = 4\), we have \(8c + 16 = 2n^2 + 4\), and hence \(n^2 = 4c + 6\).
By considering modulo 4, the only quadratic residues modulo 4 are \(0\) and \(1\), but the right-hand side equation is congruent to \(2\) modulo \(4\).
Hence, there are no solutions for \(n\) and \(c\).
When \(c = 1\), we have \(2n^2 + k = 2k + k^2\), and hence \(2n^2 = k^2 + k\).
When \(k = 1\), \(k^2 + k = 2\), and so \((n, k) = (1, 1)\) satisfies the equation.
When \(k = 8\), \(k^2 + k = 64 + 8 = 72\), and so \((n, k) = (6, 8)\) satisfies the equation.
Given that \(2N^2 = K^2 + K\), notice that \begin {align*} (2{N'}^2) - ({K'}^2 + K') & = 2 (3N + 2K + 1)^2 - (4N + 3K + 1)^2 - (4N + 3K + 1) \\ & = 2 (9N^2 + 4K^2 + 1 + 12 NK + 6N + 4K) \\ & \phantom {=} - (16N^2 + 9K^2 + 1 + 24NK + 8N + 6K) \\ & \phantom {=} - (4N + 3K + 1) \\ & = 2N^2 - K^2 - K \\ & = 2N^2 - (K^2 + K) \\ & = 2N^2 - 2N^2 \\ & = 0, \end {align*}
and this means that \[ 2{N'}^2 = {K'}^2 + K', \] and hence \[ (N', K') = (3N + 2K + 1, 4N + 3K + 1) \] is another pair of solution for \((n, k)\).
When \((n, k) = (6, 8)\), \(3n + 2k + 1 = 35\), \(4n + 3k + 1 = 49\), and \((n, k) = (35, 49)\) is also possible.
When \((n, k) = (35, 49)\), \(3n + 2k + 1 = 204, 4n + 3k + 1 = 288\), and \((n, k) = (204, 288)\) is also possible.