Find all positive integers a and b

[anemone]

Find all positive integers $a$ and $b$ such that $a(a+2)(a+8)=3^b$.

 

[lfdahl]

My solution (attempt):

Spoiler

\[(*). \;\;\;\;a(a+2)(a+8)=3^b, \: \: \: \: a,b \in \mathbb{N}\]

One obvious solution pops up for the pair: $(a,b) = (1,3)$.

(*) requires each of the three factors on the LHS to be divisible by 3:

\[(1). \;\;\;a\equiv 0\: \: \: (mod \: \: 3)\\\\ (2). \;\;\; a+2\equiv 0\: \: \: (mod \: \: 3)\\\\ (3).\; \;\; a+8\equiv 0\: \: \: (mod \: \: 3)\]

$(1)$ implies that $a = n\cdot 3$, $n \in \mathbb{N}$. This immediately excludes $(2)$ and $(3)$ for all $n$.

Hence, $a = 1$ and $b = 3$ is the only solution.

 

[kaliprasad]

Spoiler

all of a , a + 2 and a + 8 have to be power of 3( power 0 included)

a cannot pe power of 3 >= 1 or >= $3^1$ then a+ 2 and a+ 8 are not dvisible by 3

so check a = $3^0$ = 1

that give a + 2 = 3 and a + 8 = 9 both power of 3

so we get $a(a+2)(a+ 8) = 3^ 3$

so a = 1 and b= 3 is the only solution

 

[lfdahl]

My mistake. I'm sorry. Thankyou kaliprasad for your correction!

Spoiler

\[(*). \;\;\;\;a(a+2)(a+8)=3^b, \: \: \: \: a,b \in \mathbb{N}\]

One obvious solution pops up for the pair: $(a,b) = (1,3)$.

(*) requires each of the three factors on the LHS to be powers in 3:

$a = 3^n$, $n \in \mathbb{N}$. This immediately excludes $a+2$ and $a+8$ for all $n$.

Hence, $a = 1$ and $b = 3$ is the only solution.

 

[CellCoree]

After analyzing the equation, it is clear that $a$ must be a multiple of $3$ in order for the left side to be divisible by $3$. Additionally, $a+2$ and $a+8$ must also be multiples of $3$ in order for the entire equation to be divisible by $3$. This means that $a$ must be of the form $3k$, where $k$ is a positive integer.

Substituting $a=3k$ into the equation, we get $(3k)(3k+2)(3k+8)=3^b$. Simplifying, we get $27k^3+54k^2+48k=3^b$. Dividing both sides by $3$, we get $9k^3+18k^2+16k=3^{b-1}$. This shows that $3$ must be a factor of $b-1$, which means that $b$ must be of the form $3m+1$, where $m$ is a positive integer.

Substituting $b=3m+1$ into the equation, we get $9k^3+18k^2+16k=3^{3m}$. This can be rewritten as $3(3k^3+6k^2+5k)=3^{3m}$. Since $3$ is a prime number, this means that $3k^3+6k^2+5k=3^{3m-1}$. This shows that $3$ must also be a factor of $3m-1$, which means that $m$ must be of the form $3n+1$, where $n$ is a positive integer.

Substituting $m=3n+1$ into the equation, we get $3k^3+6k^2+5k=3^{3(3n+1)-1}=3^{9n+2}$. Simplifying, we get $k(3k+2)(k+1)=3^{9n+2}$. This means that $k$ must be of the form $3p$, where $p$ is a positive integer.

Substituting $k=3p$ into the equation, we get $3^9p(3p+2)(p+1)=3^{9n+2