\(% 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}}\)
Since \(q^n N = p^n\), we have \(p^n \divides q^n N\), and hence \(p \divides q^n N\).
But since \(\gcd (p, q) = 1\), we must have \(p \divides q^{n - 1} N\). Repeating this step we will get \(p \divides N\).
Let \(N = p N_1\), we have \(q^n p N_1 = p^n\), giving \(q^n N_1 = p^{n - 1}\). Repeating the same step will give \(p \divides N_1\).
Let \(N_1 = p N_2\), we have \(q^n p N_2 = p^{n - 1}\), giving \(q^n N_2 = p^{n - 2}\). Repeating the same step will give \(p \divides N_2\).
We can repeat this until we reach \(q^n N_{n - 1} = p\) from which we can conclude \(p \divides N_{n - 1}\).
So \(N_{n - 1} = kp\) for some \(k \in \NN \).
But since \(N_t = p N_{t + 1}\), we can conclude that \(N_1 = k p^{n - 1}\) and hence \[ N = p N_1 = k p^n \] as desired.
Hence, we have \(q^n k p^n = p^n\) which gives \(q^n k = 1\). B33gut this means \(q^n\) and \(k\) must both be one since \(q, k \in \NN \). Hence, \(q = 1\).
Assume, for the sake of contradiction, that \(\sqrt [n]{N}\) is a rational number that is not a positive integer. Let \[ \sqrt [n]{N} = \frac {p}{q}, \] where \(p, q \in \NN \), \(\gcd (p, q) = 1\), and \(q \neq 1\) (this is to ensure it is not a positive integer).
Hence, by rearrangement, we have \[ q^n N = p^n, \] and from what we have proved we must have \(q = 1\), which contradicts with \(q \neq 1\).
Hence, \(\sqrt [n]{N}\) must either be a positive integer or must be irrational.
Since \(a^a d^b = b^a c^b\), we know that \(a^a \divides b^a c^b\). By the same reasoning as part 1, we know that \(c^b = k a^a\) for some positive integer \(k_1\).
Hence, putting it back to the original equation, we have \[ d^b = k_1 b^a, \] which implies \(d^b \geq b^a\).
Since \(a^a d^b = b^a c^b\), we know that \(c^b \divides a^a d^b\). By the same reasoning as part 1, we know that \(a^a = k_2 c^b\) for some positive integer \(k_2\).
Hence, putting it back to the original equation, we have \[ k_2 d^b = b^a, \] which implies \(b^a \geq d^b\).
This means \(d^b = b^a\).
If a prime \(p \divides d\), then \(p \divides d^b\), and hence \(p \divides b^a\).
Since \(b^a = b b^{a - 1}\), if \(p\) does not divide \(b\), this means \(p\) and \(b\) must be co-prime (since \(p\) is a prime), then \(p\) must divide \(b^{a - 1}\), and repeating this argument eventually reaches \(p\) dividing \(b^{a - (a - 1)}\) which is a contradiction. So \(p\) must divide \(b\).
Let \(d = p^m d'\), and we must have \(p\) not divide \(d'\). Similarly, let \(b = p^n b'\), and we must have \(p\) does not divide \(b'\).
Putting this back to \(d^b = b^a\) shows \[ (p^m d')^b = (p^n b')^a, \] and hence \[ p^{mb} d'^b = p^{na} b'^a, \] and we must have \(p\) does not divide \(d'^b\) nor \(b'^a\).
This means \(p^{mb}\) and \(p^{na}\) are exactly the highest powers of \(p\) that divide \(d^b = b^a\), and hence \[ mb = na \iff b = \frac {na}{m}. \]
Since \(p^n \divides b\), we must have \(p^n \divides \frac {na}{m}\), and hence \(p^n \divides na\). However, since \(a\) and \(b\) are co-prime, and \(p\) is a prime factor of \(b\), then \(p\) must not divide \(a\), and hence \(p^n \divides n\). Hence, \(p^n \leq n\).
Since \(y^x > x\) for \(y \geq 2\) and \(x > 0\), and \(p^n \leq n\), we must have \(p < 2\) or \(n \leq 0\). But since \(p\) is a prime, \(p \geq 2\), so we must have \(n \leq 0\) and hence \(n = 0\).
This means that the highest power of the prime number \(p\) that divides \(b\) is always \(0\), and hence \(b = 1\).
Let \[ r = \frac {p}{q}, \] where \(p, q \in \NN \), \(\gcd (p, q) = 1\).
We have \[ r^r = \frac {r}{s} \] for \(r, s \in \NN \), \(\gcd (r, s) = 1\).
We have \begin {align*} \left (\frac {p}{q}\right )^{\frac {p}{q}} & = \frac {r}{s} \\ \left (\frac {p}{q}\right )^p & = \left (\frac {r}{s}\right )^q \\ p^p s^q & = q^p r^q. \end {align*}
Here, let \(a = p, b = q, c = r\) and \(d = s\). We must have \(b = q = 1\), which contradicts with \(q \neq 1\).
Therefore, \(r = p \in \NN \) is a positive integer.