Question

) If an integer is simultaneously a square and a cube,
prove that it is either of the form 7k or 7k +1. ​

Answer 1

If an integer is simultaneously a square and a cube, it will be sixth power of some integer and so it can be written in the form $\displaystyle\mathsf {a^6\ \forall a\in}\mathbb{Z}.$

Let,

$\displaystyle\sf{\longrightarrow a=7q+r}$

so that $\displaystyle\sf {a\equiv r\pmod{7}}$ assuming $\displaystyle\sf {0\leq r<7.}$

Then we get,

$\displaystyle\sf{\longrightarrow a^6\equiv r^6\pmod7}$

but $\displaystyle\sf {r^6}$ should not be $\displaystyle\sf {0\leq r^6<7}$ for every $\displaystyle\sf {0\leq r<7.}$

We need to find possible values of $\displaystyle\sf {m}$ for $\displaystyle\sf {0\leq m<7}$ such that $\displaystyle\sf{r^6\equiv m\pmod7.}$

We need to put values for $\displaystyle\sf {r}$ ranging from 0 to 6 and to check.

Let $\displaystyle\sf {r=0.}$ Then,

$\displaystyle\mathsf{\longrightarrow r^6=0^6=0\equiv}\ \mathbf{0}\ \mathsf{\pmod7}$

Let $\displaystyle\sf {r=1.}$ Then,

$\displaystyle\mathsf{\longrightarrow r^6=1^6=1\equiv}\ \mathbf{1}\ \mathsf{\pmod7}$

Let $\displaystyle\sf {r=2.}$ Then,

$\displaystyle\mathsf{\longrightarrow r^6=2^6=64\equiv}\ \mathbf{1}\ \mathsf{\pmod7\quad\quad\dots(1)}$

Let $\displaystyle\sf {r=3.}$ Then,

$\displaystyle\mathsf{\longrightarrow r^6=3^6=729\equiv}\ \mathbf{1}\ \mathsf{\pmod7\quad\quad\dots(2)}$

Let $\displaystyle\sf {r=4.}$ Then,

$\displaystyle\sf{\longrightarrow r^6=4^6=(2^2)^6=(2^6)^2}$

Using (1),

$\displaystyle\mathsf{\longrightarrow r^6=(2^6)^2\equiv1^2=}\ \mathbf{1}\ \mathsf{\pmod7}$

Let $\displaystyle\sf {r=5.}$ Then,

$\displaystyle\sf{\longrightarrow r^6=5^6=(5^3)^2}$

Since we can see $\displaystyle\sf {5^3=125\equiv-1\pmod7,}$

$\displaystyle\mathsf{\longrightarrow r^6=(5^3)^2\equiv(-1)^2=}\ \mathbf{1}\ \mathsf{\pmod7}$

Let $\displaystyle\sf {r=6.}$ Then,

$\displaystyle\sf{\longrightarrow r^6=6^6=(2\times 3)^6=2^6\times 3^6}$

Using (1) and (2),

$\displaystyle\mathsf{\longrightarrow r^6=2^6\times 3^6\equiv1\times1=}\ \mathbf{1}\ \mathsf{\pmod7}$

From each case we get,

$\displaystyle\sf{\longrightarrow m\in\{0,\ 1\}}$

so that $\displaystyle\sf {a^6}$ can only be written in the form $\displaystyle\sf{7k}$ or $\displaystyle\sf {7k+1}$ for every $\displaystyle\sf {a\in\mathbb{Z}}$ and for some $\displaystyle\sf {k\in\mathbb{Z}.}$

Hence Proved!

Answer 2

Explanation:-

$\:$

If n is both a square and a cube, it is a perfect 6-th power.

Let z be a number. Then z has shape 7k, or 7k+1, or 7k+2, and so on up to 7k+6.

Then z^6 has shape (respectively) 7k, 7k+1, 7k+1, 7k+1, 7k+1, 7k+1, and 7k+1.

To do the calculations, let us take the example z of the shape 7k+2.

Imagine expanding (7k+2)^6 using the Binomial Theorem. The first 6 terms are obviously divisible by 7, and the last term is 2^6=64. This has remainder 1 on division by 7. So (7k+2)^6 has remainder 1 on division by 7, and therefore has shape 7t+1.

The other calculations are very similar.