What is the smallest integer bigger than one that is at the same time a square and a cube and a fifth power and a seventh power?

Answer 1

Answer: 2^210.

Since 2, 3, 5 and 7 are prime, the smallest common multiple equals 2x3x5x7 = 210. We are looking for an x with the property:

x = b^2 =c^3 = d^5 = e^7,

hence, e needs to be a square, a cube and a fifth power. This gives us a similar problem with one less constraint.

e = f^2 = g^3 = h^5

hence h needs to be a square and cube. This gives us

h = j^2 = k^3

hence k needs to be a square, so

k = a^2 for some a, and

h = k^3 = (a^2)^3 = a^6, and

e = h^5 = a^30, and

x = e^7 = a^210

The smallest integer >1 is obtained for a=2.

Note that 2^210 ~ 1.64 x 10^63 and is so large that the chances to find this number anywhere in nature are limited. For instance, the Earth has only about 10^46 atoms.