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

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

• You started by saying "Suppose $n^2- 2$ is divisible by 4". What about the other cases where $n^2- 2$ is not divisible by 4? Nov 24 '18 at 13:00
• sorry i meant Suppose $n^2−7$ is divisible by $5$ Nov 24 '18 at 13:37

## 6 Answers

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.

• sorry i meant Suppose n2−7 is divisible by 5 Nov 24 '18 at 18:35
• I know, and the above precisely proves that there is no $n$ with that property. Nov 24 '18 at 18:37

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 $$5$$.

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.

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?

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, $$n^2-2\neq 0(mod5)$$.

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 $$5$$.