Prove that $7n^3-1$ can never be a perfect square.

Question

Prove that $7n^3-1$ can never be a perfect square.

hint: consider about congruences concepts.

I noticed that $7n^3$ is congruent to 6 mod 7 but I don't know if there is any definition or theorem about perfect square I can use in this question?

Answer 1

Suppose it can be a square. Then there exist natural number $k$ such that $$7n^3=k^2+1$$

So $7|k^2+1$. There is a theorem that states that if $p\equiv 3 \pmod 4$ and $p|a^2+b^2$ that $p|a$ and $p|b$. So it is impossible.

Answer 2

Let us go through a list of squares numbers through modulo 7. It is a somewhat 'crude' method but it works nonetheless.

\begin{align} 1^2 &\equiv 1\pmod 7 \\ 2^2 &\equiv 4\pmod 7 \\ 3^2 &\equiv 2\pmod 7 \\ 4^2 &\equiv 2\pmod 7 \\ 5^2 &\equiv 4\pmod 7 \\ 6^2 &\equiv 1\pmod 7 \\ 7^2 &\equiv 0\pmod 7 \\ 8^2 &\equiv 1\pmod 7 \\ 9^2 &\equiv 4\pmod 7 \\ 10^2 &\equiv 2\pmod 7 \\ 11^2 &\equiv 2\pmod7\\ & \space\space\space\space\space\space\space\vdots \end{align}

Notice that the pattern repeats every 7 terms. Also, notice that there is no number $n^2$ such that $n^2 \equiv 6 \pmod 7$.

You rightly found that $7n^3−1 \equiv 6\pmod 7$. That means $7n^3−1$ cannot equal $n^2$ for any $n \in \mathbb Z$.

$\therefore$ $7n^3−1$ is not a perfect square.

---

As for theorems on this topic, it is important to know the following: $$a \equiv b \pmod m$$ $$\iff m|a-b$$

Furthermore, if $$a \equiv b \pmod m$$ and $$c \equiv d \pmod m$$

then the following theorems apply.

$$a \equiv c \pmod m$$ $$a+c\equiv b+d \pmod m$$ $$a \times c \equiv b \times d \pmod m$$