Cubist in the City

Positive prime numbers start from $2,3,5,7......$ .

For $p = 2,$

$p^{14} + 55p^{12}$

$=2^{14} + 55 \times 2^{12}$

$= 241664 \longrightarrow \text{Not a perfect cube}$

For, $p = 3,$

$p^{14} + 55p^{12}$

$=3^{14} + 55 \times 3^{12}$

$= 34012224 \longrightarrow \text{Perfect cube}$

Hence $p= 3$ is smallest.

Given that p^14 + 55p^12 = k^3, k= n p^4 for some positive integer n. Then p^14 + 55p^12 = (n^3) p^12. Dividing by p^12, p^2 + 55 = n^3. Clearly, p=3 and n = 4. Ed Gray

We notice that, $p^{12}=(p^4)^3$ is a perfect cube. So, we'll factorize our expression a bit:

$p^{14}+55p^{12}=p^{12}(p^2+55)$

Now we happily declare $(p^2+55)$ to be a perfect cube. Since $p^2\geq4$ , the inequality $27=3^3 <59\leq p^2+55$ holds. Therefore, for smallest $p$ , $(p^2+55)$ is $4^3=64$ .

Solving $p^2+55=64$ returns the value $\boxed3$ as our intended answer.

$Q_t\pi$