If $n$ is a perfect square and a perfect cube then $7$ divides $n(n-1) $

Question

If a positive integer $n$ is both a perfect square and a perfect cube , then is it true that $7$ divides $n(n-1)$ ?

Answer 1

The result is true:

Since $n$ is a perfect square and a perfect cube, $n=m^6$ for some $m$.

Now, if $m$ is divisible by $7$ then we are done.

Otherwise, note that $m^6-1=(m^3-1)(m^3+1)$. $m$ can only leave remainders $\pm1,\pm2,\pm3$ when divided by $7$. In all these cases, it's easily checked that either $m^3-1$ or $m^3+1$ leaves remainder $0$ when divided by $7$. (I am implicitly using the language of congruence modulo $7$).

Answer 2

For any such number $n$, there exists an $x \in \mathbb{Z}$ such that $n = x^6$.

From here, we know $x$ is equivalent to $0, 1, 2, 3, 4, 5$, or $6 \pmod{7}$.

If you check each of these, you will find that, for all $x \not\equiv 0 \pmod{7}$, $x^6 \equiv 1 \pmod{7}$.

And so...

Answer 3

You can write $n=m^6$. Then, you use the Euler's theorem. If $m=0\mod 7$, then $$n=0 \mod 7$$ and the first factor in $n(n-1) \mod 7$ is $0$. In all other cases, $m$ and $7$ are coprime and by Euler's theorem, it holds $$n=m^{7-1}=1\mod 7$$ meaning that $$n-1=0\mod 7$$ Both results together mean that $n(n-1)=0\mod 7$.

Answer 4

Hints: Start from the observation that $7$ is a prime number.

For every prime number $\color{red}{\mathbf p}$ and every integer $a$ not a multiple of $\color{red}{\mathbf p}$, by a very famous theorem you surely know, $a^{\color{red}{\mathbf p}-1}$ equals $____$ modulo $\color{red}{\mathbf p}$. Hence, for every integer $a$ and every positive $\color{green}{\mathbf k}$, $$a^\color{green}{\mathbf k}(a^{\color{red}{\mathbf p}-1}-1)=\text{____}\pmod{\color{red}{\mathbf p}}, $$ since, either $a$ is a multiple of $\color{red}{\mathbf p}$ and then the result is obvious, or $a$ is not a multiple of $\color{red}{\mathbf p}$ and then we apply the "very famous theorem" used above.

How to apply these facts? Well, the integer $n$ in your question is a sixth power $n=a^\color{purple}{\mathbf 6}$ and you are interested in the divisibility of $n(n-1)=a^\color{green}{\mathbf6}(a^\color{purple}{\mathbf 6}-1)$ by $\color{red}{\mathbf p}=7$. This $\color{red}{\mathbf p}$ is prime and such that $\color{red}{\mathbf p}-1=\color{purple}{\mathbf 6}$. Does this connect the dots?

Answer 5

Let's solidify the argument by showing that if $x = n^2 = m^3$ then $x = k^6$ for some $k \in \mathbb{N}$, then after this is done, then we can use either of the above answers. Write $n = p_1^{r_1}p_2^{r_1}\cdot...p_t^{r_t}$, and $m = q_1^{s_1}q_2^{s_2}\cdot...q_d^{s_d}$, then: $n^2 = m^3 \to p_1^{2r_1}p_2^{2r_2}\cdot...p_t^{2r_t} = q_1^{3s_1}q_2^{3s_2}\cdot...q_d^{3s_d}$, with $p_i, q_j$ are prime numbers, and $p_i \neq p_j$ if $i \neq j$, and $q_i \neq q_j$ if $i \neq j$. This equation also shows that $p_i = q_j$ for all $i = 1,2,..,t$. So: $t = d$, and $2r_i = 3s_j$ for all $i = 1,2, ...t$. But this means: $3 | r_i$ for all $i$'s. So $r_i = 3w_i$, and $2r_i = 6w_i$ for all $i$'s. Thus: $x = p_1^{6w_1}p_2^{6w_2}\cdot...p_t^{6w_t} = \left(p_1^{w_1}p_2^{w_2}\cdot...p_t^{w_t}\right)^6 = k^6$, with $k = p_1^{w_1}p_2^{w_2}\cdot...p_t^{w_t}$

Answer 6

The quadratic residues mod $7$ are $\{0,1,2,4\}$. The cubic residues mod $7$ are $\{0,1,6\}$. Thus, if $n$ is a square and a cube, it must be either $0$ or $1$ mod $7$. Thus, $n(n-1)\equiv0\pmod{7}$.