Squares over, now are cubes!

Case 1: $n<9$

First we show that $3^9+3^{12}+3^{15}+3^n$ fails to be a perfect cube if $n<9$ . Using the binomial expansion, we see that $(3^5+3)^3<3^9+3^{12}+3^{15}+3^n$ $<(3^5+4)^3$ for all $n<9$ . Thus $3^9+3^{12}+3^{15}+3^n$ is "trapped" between two consecutive perfect cubes, $246^3$ and $247^3$ .

Case 2: $n \geq9$

We observe that $3^9+3^{12}+3^{15}+3^n=3^9(1+3^3+3^6+3^{n-9})$ $=27^3(757+3^{n-9}).$ We want $757+3^{n-9}$ to be a perfect cube. Now the smallest perfect cube exceeding $757$ is $1000$ , and it so happens that $1000-757$ $=243=3^5.$ Thus $3^9(1+3^3+3^6+3^5)$ $=3^9+3^{12}+3^{15}+3^{14}=270^3$ .

The answer is $n=14$ .

$3^9+3^{12}+3^{15} = 3^9(1+27+729) = 3^9\cdot 757.$ The closest cubes are $9^3=729,$ and $10^3=1000$ . Since $1000-757=243=3^5, n=9+5=\boxed{14}.$

• We can rearrange this expression as $3^{9}(1 + 3^{3} + 3^{6} + 3^{n - 9}) = 3^{9}(757 * 3^{n - 9})$ .
• Since $3^{9}$ is a perfect cube, we don't have to worry about it anymore.
• We also can see that $\frac{757}{3}$ has 3 as remainder, so does $\frac{1000}{3}$ (who is actually the closest perfect cube with 3 as remainder).
• Giving a shot, we try $757 + 3^{n - 9} = 1000 = 10^{3}$ $3^{n - 9} = 243 = 3^{5}$ $n - 9 = 5 => n = 14$ .
• So, we conclude that the minumum possible value of positiver integer n such that the expression in question is a perfect cube is 14 .

The first thing I noticed was that $3^n$ will be odd for any positive integer n so the sum of 4 of these must be even. As the number is a cube, it must be the product of cube primes. As 2 divides this number, 8 must divide it too. Also, it is the sum of multiples of 3 so the final number must be divisible by 3 cubed or 27. As the terms are multiples of 27, $3^n$ must be too so $n \geq 3$

In order for the number to be a multiple of 8, consideration is needed for the terms mod 8. $3^9 \equiv 3 \mod 8 \\ 3^{12} \equiv 1 \mod 8 \\ 3^{15} \equiv 3 \mod 8 \\ \implies 3^n \equiv 1 \mod 8$ Which shows that n has to be even.

At this point. I started getting lazy so I made a program in VB.net that generates these numbers and I put in that n has to be even as to save time. Luckily, I was able to find $\boxed{n=14}$ . Sorry about the anticlimactic ending :(.