electrovector wrote:

Can you show your solution in detail please

@above

Q.E.D$\blacksquare$

I think I have done something wrong while using hide codes Is there anyone to help me

@above Can you explain why when $p=19$, the only possible solutions of $2a+11$ are ${3(19^n),19^n,3(19^{n-1}),57}$?

Because of gcd and $2a+11>a-4$. Just subcasing. Bruh...

My Solution: Let $d=gcd(2a+11,a-4)\Rightarrow d\mid 2a+11$ and $d\mid a-4\Rightarrow d\mid 2a-8\Rightarrow d\mid 2a-2a+11+8\Rightarrow d=1,19$ So we have two cases: Case 1:$gcd(2a+11,a-4)=1$ $\Rightarrow p^{n}\mid 2a+11$ or $p^{n}\mid a-4$ If $p^{n}\mid 2a+11\Rightarrow \frac{3}{a-4}\in N\Rightarrow a=5,7$ If $p^{n}\mid a-4\Rightarrow \frac{3}{2a+11}\in N$ which is impossible. Plug in $5$ and $7$ instead of $a$ and we get that our triples are:$(a,n,p)=(5,1,7);(7,2,5)$ Case 2:$gcd(2a+11,a-4)=19$ $\Rightarrow 2a+11=19k,a-4=19l;gcd(k,l)=1\Rightarrow 361kl\mid 3p^{n}\Rightarrow 361\mid 3p^{n}\Rightarrow 361\mid p^{n}\Rightarrow p=19$ $\Rightarrow 361kl=3\times 19^{n}$ Suppose that $19^{n}\mid kl\Rightarrow kl=19^{n}\times m;m\in Z^{+}\Rightarrow m=\frac{3}{361}$,but we know that $m\in Z^{+}$,contradiction. And since $19^{n}\mid 361kl$ and $19^{n}\nmid kl$ we have $19^{n}\mid 361$ $19^{n}\mid 361$ , $361\mid 19^{n}\Rightarrow 361=19^{n}\Rightarrow n=2\Rightarrow a=23$ So our final solutions are:$(a,n,p)=(5,1,7);(7,2,5);(23,2,19)$ $Q.E.D$

electrovector wrote: I think I have done something wrong while using hide codes Is there anyone to help me I think it's supposed to be "[/hide]", not "[\hide]".

Kaisun wrote: electrovector wrote: I think I have done something wrong while using hide codes Is there anyone to help me I think it's supposed to be "[/hide]", not "[\hide]". Didn't work. Doesn't matter.

Redacted.

You can also solve this problem like I did haha: Just take discriminant of given equation and solve a diophantine equation: $24p^{n}=(x-19)(x+19)$, where $x$ is positive integer. There are many cases, but it does not take a while to solve like this also.

electrovector wrote: Solve $2a^2+3a-44=3p^n$ in positive integers where $p$ is a prime. $(a-4)(2a+11)=3p^n$ Case 1-: let $3|2a+11\implies a\equiv 2\mod 3$ Let $a=3k+2$ then we have $(3k-2)(2k+5)=p^n$ so $3*(2k+5) -2(3k-2)=3*p^{n-x} - 2p^x \implies 19=p^x(3*p^{n-2x} -2)$ hence $p=19, x=1, n=2$ hence $k=7\implies a=23$ so $(p, a, n)=(19,23,2)$ in this case. Case 2-: let $3|a-4\implies a\equiv 1\mod 3$ let $a=3k+1$ then we have $(k-1)(6k+13)=p^x$ So $(6k+13) -6*(k-1) =p^{n-x} -6*p^x \implies 19=p^x(p^{n-2x}-6)$ clearly $p=19, x=1$ has no solution for this case. Case 3-: $a-4=3$ or $2a+11=3$ but $2a+11>3$ so $a-4=3,a=7$ so $2a+11=25=5^2$ so in this case we have $(p, a, n)=(5, 7,2)$ Case 4-: $a-4=1$ or $2a+11=1$ but $2a+11>1$ so $a-4=1, a=5\implies 2a+11=21=3*7^1$ Hence in this case we have $(p, a, n)=(7, 5,1)$

$$2a^2+3a-44=3p^n$$$$\implies (2a + 11)(a-4) = 3p^n$$Notice that $\gcd(2a + 11 , a-4) = \gcd(a-4 , 2a + 11 - 2a + 8) = \gcd(a-4 , 19) \in \{1,19\}$ Case 1 : Let $\gcd = 1$ Thus $3$ will contribute in one factor and $p^n$ on other or else the other possibility is $3p^n$ will contribute to one factor and other factor will be equal to $1$ Since $2a + 11 > a-4$ Hence $a-4 = 1$ or $a-4 = 3$ From here we get $(a,p,n) = (5,7,1), (7,5,2)$ Case 2 : $\gcd = 19$ $\implies 19 \mid LHS \implies 19 \mid RHS \implies 19 \mid p^n \implies p = 19$ $\implies (2a + 11)(a-4) = 3\cdot 19^n$ Subcase 1 : $2a + 11 = 3 \cdot 19^m \quad \text{and} \quad a-4 = 19^{n-m}$ Notice that if both $m , n-m > 1 \ge 2 \implies \gcd > 19$ which is a contradiction. Hence one of them has to equal $1$. When $m=1$ We get $a = 28 \implies a-4 = 24 \neq 19^{n-m}$, hence no solution. When $n-m = 1$ We get $a - 4 = 19 \implies a = 23$ Which works and we get another triple $(a,p,n) = (23 , 19 , 2)$ Subcase 2 : $2a + 11 = 19^m \quad \text{and} \quad a-4 = 3 \cdot 19^{n-m}$ When $m=1$ We get $a = 4 \implies a-4 = 0$, hence no solution. When $n-m = 1$ We get $a = 61 \implies 2a + 11 = 133 \neq 19^{m}$, hence no solution. So our solutions are $\boxed{(a,p,n) = (5,7,1), (7,5,2) , (23 , 19 , 2)}$

2022 high school first round Last Number theory question a bit easy didnt think the solution in exam

A really similar problem appeared in India at a computational contest. PRMO 2017