\(% 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}}\)

2018.2.6 Question 6

1. Notice that for \(n \geq 5\), \(n! = 5 \cdot 4! \cdot 6 \cdot 7 \cdots n\), and \(n! = 5k\) for \(k = 4! \cdot 6 \cdot 7 \cdots n > 1\) is an integer.

   Therefore, \[ n! + 5 = 5k + 5 = 5 (k + 1) \] is a multiple of two integers greater than \(1\), and hence \(p\) cannot be prime.

   Hence, \(n < 5\).

   If \(n = 1\), \(n! + 5 = 6\) is not prime.

   If \(n = 2\), \(n! + 5 = 7\) is prime. \((n, p) = (2, 7)\) is a solution.

   If \(n = 3\), \(n! + 5 = 11\) is prime. \((n, p) = (3, 11)\) is a solution.

   If \(n = 4\), \(n! + 5 = 29\) is prime. \((n, p) = (4, 29)\) is a solution.

   Therefore, all solutions are \((n, p) = (2, 7), (3, 11)\) and \((4, 29)\).

2. If \(n \geq 7\), then we have \[ m! = 1! \times 3! \times \cdots \times (2n - 1)! > (4n)! \] and hence \(m > 4n\).

   Let \(p\) be some prime number between \(2n\) and \(4n\). Therefore, \(m!\) must include \(p\) as one of its terms, and \(p \divides m! = \RHS \).

   However, on the left-hand side, all the terms are less than \(p\), and since \(p\) is a prime, it must not divide any term in the left-hand side factorial expansion (since every term in the expansion is less than \(p\)), and hence \(p \notdivides \LHS \).

   But since \(\LHS = \RHS \) this is impossible, and we can deduce that \(n < 7\).

   • \(n = 1\), \(\LHS = 1! = 1\) and \((n, m) = (1, 1)\) is a solution.
   • \(n = 2\), \(\LHS = 1! \cdot 3! = 3!\) and \((n, m) = (2, 3)\) is a solution.
   • \(n = 3\), \(\LHS = 1! \cdot 3! \cdot 5! = 6 \cdot 5! = 6!\) and \((n, m) = (3, 6)\) is a solution.
   • \(n = 4\), \(\LHS = 1! \cdot 3! \cdot 5! \cdot 7! = 6! \cdot 7! = 7! \cdot 6! = 7! \cdot (3 \cdot 2 \cdot 5 \cdot 4 \cdot 3 \cdot 2) = 7! \cdot (2 \cdot 4) \cdot (3 \cdot 3) \cdot (2 \cdot 5) = 10!\) and \((n, m) = (4, 10)\) is a solution.
   • \(n = 5\), \(\LHS = 1! \cdot 3! \cdot 5! \cdot 7! \cdot 9! = 10! \cdot 9! > 10!\), so if \(m\) exists, \(m > 10\) and \(m \geq 11\). Then \(11 \divides \RHS = \LHS \), but this is impossible since \(11 > 9\), so such \(m\) does not exist.
   • \(n = 6\), \(\LHS = 1! \cdot 3! \cdot 5! \cdot 7! \cdot 9! \cdot 11! = 10! \cdot 9! \cdot 11! = 11! \cdot 9! \cdot 10! = 12! \cdot 10! \cdot (9 \cdot 8 \cdot 7 \cdot 5 \cdot 4 \cdot 3) > 12!\), so if \(m\) exists, \(m > 12\) and \(m \geq 13\). Then \(13 \divides \RHS = \LHS \), but this is impossible since \(13 > 11\), and so such \(m\) does not exist.

   Hence, the only possible solutions are \[ (n, m) \in \{(1, 1), (2, 3), (3, 6), (4, 10)\}. \]