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What are the integer values that

$$\frac {8 + 4 \cdot 3^{a}+2 \cdot 3^{a+b}+3^{a+b+c}} {3^{a+b+c+d} - 16}$$

can take when $a, b, c, d$ are positive non-zero integers?

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It's easy to prove the bound $2\cdot 3^{a+b}<3^{a+b+c}$ and the same telescopically for the rest of the terms, so the numerator $<4\cdot 3^{a+b+c}$. Given that $a,b,c,d$ are nonzero positives, the largest upper bound you can possibly imagine is when $a=b=c=d=1$: you have $<4\cdot 3^3/(3^4-16)\approx 1.66$. So the only possible integer value of this expression is $1$ (it can't be zero or negative, the denominator can't be below 65).

Now you have an equation to solve: $$8+4\cdot 3^a + 2\cdot 3^{a+b}+3^{a+b+c}=3^{a+b+c+d}-16$$

$$24+3^a(4 + 3^b(2+3^c(1-3^d))=0$$ $$3^{a-1}(4 + 3^b(2+3^c(1-3^d))=-8$$ Factorization requires $a=1$. Proceed: $$3^b(2+3^c(1-3^d))=-12$$ Factorization gives $b=1$. Proceed: $$3^c(1-3^d)=-6$$ Again, $c=1$. $$-3^d=-3$$ So $d=1$. This cascade showed that the only solution that gives an integer, is $a=b=c=d=1$, in which case the result is $1$. The estimate used in the first paragraph was an overestimation, because I said $8<3^{a+b+c}$, $4<3^{b+c}$ and $2<3^{c}$ due to $a,b,c,d$ being equal or greater than one. This also tells us... the result of the expression is in the interval (0,1].

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You need the numerator be a multiple of the successive $3^n-16$ for $n\ge 4$ of the denominator.

For the minimun value $n=4$ the numerator is univocally determined, given the fraction $\frac ND=\frac{65}{65}=1$ This is the solution $(a,b,c,d)=(1,1,1,1)$

For $n=5$ we get $D=227$ and the greatest numerator $N=195$ so there is no solution for $n=5$.

You can verify the same $N<D$ for greater values of $n$.

Thus there is just the one solution above.

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