Proof: $n^2 - 7$ is not divisble by 5

Question

I tried to prove that $n^2 - 7$ is not divisible by $5$ via proof by contradiction. Does this look right?

Suppose $n^2 - 7$ is divisible by $5$. Then:

$n^2 - 7 = 5g$, $g \in \mathbb{Z}$.
$n^2 = 5g + 7$.

Consider the case where $n$ is even.

$(2x)^2 = 5g + 7$, $x \in \mathbb{Z}$.

$4x^2 = 5g + 7$.

$4s = 5g + 7$, $s = x^2, s \in \mathbb{Z}$ as integers are closed under multiplication.

$2s$ is even, and $5g + 7$ is odd if we consider that g is an even number. so we have a contradiction.

Consider the case where $n$ is odd.

$(2x + 1)^2 = 5g + 7$, $x \in \mathbb{Z}$

$4x^2 + 4x = 5g + 6$

$4x^2 + 4x = 5g + 6$

$4(x^2 + x) = 5g + 6$

$4j = 5g + 6$, $j = x^2 + x, j \in \mathbb{Z}$ as integers are closed under addition

$2d = 5g + 7$, $d = 2j; d \in \mathbb{Z}$ as integers are closed under multiplication

$2d$ is even, and $5g + 7$ is odd if we consider that g is an odd number. An even number cannot equal an odd number, so we have a contradiction.

As both cases have a contradiction, the original supposition is false, and $n^2 - 7$ is not divisible by $5$.

is my proof correct because I cannot prove that $5g + 6$ or $5g + 7$ are odd so I assumed that in the even case g is even and in the odd case g is odd

Answer 1

It seems rather overcomplicated. Instead, simply note that the question is equivalent to solving $n^2 - 2 = 0$ in $\mathbb{Z}/5\mathbb{Z}$. But there are only five values in $\mathbb{Z}/5\mathbb{Z}$, and we can just try them all to see if they're solutions:

\begin{align*} 0^2 - 2 &= -2 \neq 0\\ 1^2 - 2 &= -1 \neq 0\\ 2^2 - 2 &= 2 \neq 0\\ (-1)^2-2 &= -1 \neq 0\\ (-2)^2-2 &= 2 \neq 0. \end{align*}

None of them are solutions, so there are no solutions to the original problem.

Answer 2

Here is an alternate approach. $n^2-7$ is divisible by 5 only if the last digit of $n^2-7$ is either 0 or 5. This means the last digit of $n^2$ is either 7 or 2. Now can you complete the proof?

Answer 3

Want to show $n^2-7 \neq 0 (mod 5)$ which is the same as show $n^2-2 \neq 0 (mod5)$. Here $0^2=0 (mod5), 1^2=1 (mod5), 2^2=4 (mod5), 3^2=4 (mod5), 4^2=1 (mod5)$. Then, that means $n^2=1, 4 (mod5)$. SO if it is $1$, then $n^2-2=-1(mod5)$ and if $n^2=4$, $n^2-1=3(mod5)$. In all cases, $n^2-2\neq 0(mod5)$.

Answer 4

Suppose $n^2-7$ is divisible by $5$ i.e. $n^2\equiv 7 \mod 5\implies n^2\equiv 2\mod 5$. We can check $\:\forall n\in\mathbb{Z}\quad$ $$n^2\equiv\begin{cases}0 &\mod5\\1 &\mod5\\4 &\mod5\end{cases} $$

So $n^2-7$ is not divisible by $5$.

Answer 5

If possible let $ n^2-7$ is divisible by $5$, then we have

$n^2-7=5k, \ \ k \in \mathbb{Z}, ........(1)$.

Now let $ k \in \mathbb{Z}^{+}$, then for $N ( \neq 7) \in \mathbb{N}$, we have $k=n=N$ ( without loss of generality)

Then from $(1)$, we get

$ N^2-7=5N \\ \Rightarrow N^2-5N-7=0 \\ \Rightarrow N=7, -2$

But as $N \neq 7$, we must have $ N=-2 \in \mathbb{N}$=set of natural number, which is impossible.

Thus $n^2-7$ is not divisible by $5$.

Answer 6

Your proof is incorrect. Note $\,n^2 = 5g+7 \,\Rightarrow\bmod 2\!:\ n\equiv g+1\,$ so $n$ and $g$ have opposite parity. But your proof only considers the cases when they have equal parity.

Instead note $\bmod 5\!:\ n\equiv 0,\pm1,\pm2\,\Rightarrow\,n^2\equiv 0,1,4 \not\equiv 7$

Alternatively $\,n^2\equiv 7\equiv 2\,\Rightarrow\, n^4\equiv 2^2\not\equiv 1\,$ contra little Fermat.