\[|(2n+3)(3n-13)| = p^k\] $gcd(2n+3,3n-13) = gcd(2n+3, 6n-26) = gcd(2n+3,35) = 1,5,7,35$ so finitely many $n$ to check. if $gcd(2n+3,3n-13) = 1$, then either $n=-1$ or $n=4$ (one of the factors has to be one). This gives $(n,k,p) = (4,1,11)$. if $gcd(2n+3,3n-13) = 5$, then one has to equal $\pm 5$. this gives $n=\{-4,1,6\}$ and u get $(n,k,p) = (-4,3,5)$ if $gcd(2n+3,3n-13) = 7$, then one has to equal $\pm 7$. this gives $n = \{-5,2\}$ and u get $(n,k,p) = (2,2,7)$ $gcd =35$ not possible because then $35|p$, contradiction. Therefore we're done

straight wrote: \[|(2n+3)(3n-13)| = p^k\] $gcd(2n+3,3n-13) = gcd(2n+3, 6n-26) = gcd(2n+3,35) = 1,5,7,35$ so finitely many $n$ to check. if $gcd(2n+3,3n-13) = 1$, then either $n=-1$ or $n=4$ (one of the factors has to be one). This gives $(n,k,p) = (4,1,11)$. if $gcd(2n+3,3n-13) = 5$, then one has to equal $\pm 5$. this gives $n=\{-4,1,6\}$ and u get $(n,k,p) = (-4,3,5)$ if $gcd(2n+3,3n-13) = 7$, then one has to equal $\pm 7$. this gives $n = \{-5,2\}$ and u get $(n,k,p) = (2,2,7)$ $gcd =35$ not possible because then $35|p$, contradiction. Therefore we're done You forgot $p = 19, n = -2, k = 1$

My solution during the contest. Answer:$(-2,1,19),(-1,4,2),(4,1,11),(2,2,7),(-4,3,5)$. Firstly $k \geq 0$. $|(2n+3)(3n-13)|=p^k$ $|2n+3|=1 \implies n=-2,-1 \implies (-2,1,19),(-1,4,2)$ $|3n-13|=1 \implies n=4 \implies (4,1,11)$ $|2n+3|=p^x,|3n-13|=p^y$ such that $x,y>0$. $i)x>y$ $p^y=3n-13|2n+3 \implies 3n-13|3(2n+3)-2(3n-13)=35$ $|3n-13|=5,7 \implies n=6,2 \implies (2,2,7)$ $ii)x \leq y$ $p^x=2n+3|3n-13 \implies 2n+3|3(2n+3)-2(3n-13)=35$ $|2n+3|=5,7 \implies n=-4,1,-5,2 \implies (-4,3,5),(2,2,7)$

we have $|2n+3|\cdot |3n-13|=p^{k}$. now clearly we must have $k \geqslant 0$ Consider $|2n+3|=p^{\alpha} , |3n-13|=p^{\beta}$ , where $\alpha , \beta \in \mathbb{Z}_{0}^{+}$ clearly we can't have $\alpha=\beta=0$, so: Case 1:- $\alpha=0$ then we have $|2n+3|=1$ which gives $n=-1$ or $-2$, plugging back into the given relation we observe they indeed work for $p=2, k=4$ and $p=19$ and $k=1$ respectively. Case 2:- if $\beta$=0 we get $|3n-13|=1$ , which gives $n=4$ to be the only possibility, plugging back we get $p=11$ and $k=1$ Now if $\alpha , \beta \in \mathbb{Z}^{+}$ we observe: Case 3:- $\alpha \geqslant \beta$ we observe $(|3n-13|)|(|2n+3|) \iff 3n-13|2n+3 \iff 3n-13| 3(2n+3)-2(3n-13) \iff 3n-13|35$ which gives $n=2$ is the only workable number which gives $p=7$ and $k=2$. Case 4:- $\beta >\alpha$ we proceed similarly as Case 3 to get: $2n+3|35$ which gives $n=-4$ is the only workable number which gives $p=5$ and $k=3$ In conclusion the solution set is: $\boxed{\{(n,p,k):(-1,2,4),(-2,19,1),(4,11,1),(2,7,2),(-4,5,3)\}}$ $\blacksquare$

bin_sherlo wrote: My solution during the contest. Answer:$(-2,1,19),(-1,4,2),(4,1,11),(2,2,7),(-4,3,5)$. Firstly $k \geq 0$. $|(2n+3)(3n-13)|=p^k$ $|2n+3|=1 \implies n=-2,-1 \implies (-2,1,19),(-1,4,2)$ $|3n-13|=1 \implies n=4 \implies (4,1,11)$ $|2n+3|=p^x,|3n-13|=p^y$ such that $x,y>0$. $i)x>y$ $p^y=3n-13|2n+3 \implies 3n-13|3(2n+3)-2(3n-13)=35$ $|3n-13|=5,7 \implies n=6,2 \implies (2,2,7)$ $ii)x \leq y$ $p^x=2n+3|3n-13 \implies 2n+3|3(2n+3)-2(3n-13)=35$ $|2n+3|=5,7 \implies n=-4,1,-5,2 \implies (-4,3,5),(2,2,7)$ FOR CASE k<0?

kundethuong2k7 wrote: bin_sherlo wrote: My solution during the contest. Answer:$(-2,1,19),(-1,4,2),(4,1,11),(2,2,7),(-4,3,5)$. Firstly $k \geq 0$. $|(2n+3)(3n-13)|=p^k$ $|2n+3|=1 \implies n=-2,-1 \implies (-2,1,19),(-1,4,2)$ $|3n-13|=1 \implies n=4 \implies (4,1,11)$ $|2n+3|=p^x,|3n-13|=p^y$ such that $x,y>0$. $i)x>y$ $p^y=3n-13|2n+3 \implies 3n-13|3(2n+3)-2(3n-13)=35$ $|3n-13|=5,7 \implies n=6,2 \implies (2,2,7)$ $ii)x \leq y$ $p^x=2n+3|3n-13 \implies 2n+3|3(2n+3)-2(3n-13)=35$ $|2n+3|=5,7 \implies n=-4,1,-5,2 \implies (-4,3,5),(2,2,7)$ FOR CASE k<0? If $k<0$, then $LHS$ is an integer and $RHS$ is not?