Question

Write cubes of $5$ natural numbers which are of the form $3n+1$ (e.g. 4,7,10,....) and verify the following:
'The cube of a natural number of the form $3n+1$ is a natural number of the same form'.

Answer 1

The 5 natural number which are of the form $3n+1$ (e.g 4,7,10,....) are as follows:
$3\times 1+1=3+1=4$
$3\times 2+1=6+1=7$
$3\times 3+1=9+1=10$
$3\times 4+1=12+1=13$
$3\times 5+1=15+1=16$
The cubes of 5 natural numbers which are of the form $3n+1$ (e.g. 4,7,10,....) are as follows:
${\left(4\right)}^3=4\times 4\times 4=64$
${\left(7\right)}^3=7\times 7\times 7=343$
${\left(10\right)}^3=10\times 10\times 10=1000$
${\left(13\right)}^3=13\times 13\times 13=2197$
${\left(16\right)}^3=16\times 16\times 16=4096$
Verification:
$64=3\times 21+1$
$343=3\times 114+1$
$1000=3\times 333+1$
$2197=3\times 732+1$
$4096=3\times 1365+1$
$\therefore$ 'The cube of a natural number of the form $3n+1$ is a natural number of the same form'