Suppose that n is an integer such that 5|(n + 2)

Question

Suppose that n is an integer such that $5|(n + 2)$. Which of the following are divisible by $5$?

$n^2 - 4$, $n^2 + 8n + 7$, $n^4 - 1$, $n^2 - 2n$

Answer 1

Please avoid asking solutions to problems unless they are research oriented and worthy of Math Stack Exchange Veterans (I'm obviously not one of em lol). Stack Exchange ain't not a place to get specific problems solves. I recommend Art Of Problem Solving for Questions like these.

Rather go for something like: "Approaching Divisibility Problems". "I have a divisibility statement. How do I proceed to find if other expressions involving the same variable follow this divisibility too?"

If I had enough Reputation I'd vote to close your question right now.

If your question was the one I mentioned in the previous paragraph, Then here you go:

Learn Modular Algebra if you have enough time. It deals with divisibility and similar problems. $n=3mod5$ If this scares you just take $n = 5k - 2$ and proceed.

Don't ask solutions to problems in stackexchange.

Answer 2

$n+2 \equiv 0 \pmod 5 \implies n \equiv -2 \pmod 5$.

Factoring all the expressions makes it easier, even though it is not strictly necessary.

The first one is shown as an example:

$n^2 - 4 = (n+2)(n-2) \equiv (0)(-4) = 0 \pmod 5$, so it is divisible by $5$.

Working this way, you'll find every single one has a factor that is $0$ or a multiple of $5$ (which is equivalent to $0 \pmod 5$) except for the last one.

So the answer is all except the last, namely $n^2 - 2n = n(n-2)$.

Answer 3

Here is an approach without explicit modular arithmetic, just basic algebra.

$5|(n + 2)$ means $\dfrac{(n+2)}{5} = k$ where $k$ is an integer. Therefore:

$$5|(n + 2) \implies \dfrac{(n+2)}{5} = k \implies (n+2) = 5k \qquad k \in \mathbb{Z} $$

In other words, this means that we can replace $(n+2)$ with $5k$.

Now, recall that any number that can be written as a multiple of $5$ is itself divisible by $5$. Therefore, your goal is to rewrite the expressions in such a way that you have the $n+2$ terms isolated.

For instance, look at the first expression in the list. Using factorisation, you get the following:

$$n^2 - 4 = (n+2)(n-2)$$

Now, there is an isolated $(n+2)$ term. Therefore, we can replace it with $5k$ as explained. This gives:

$$n^2 - 4 = (n+2)(n-2) = 5k(n-2)= 5\cdot k(n-2)$$

From here it should be clear that the resultant expression is a multiple of $5$ therefore, it is divisible by 5.

Perform similar manipulations to the rest of the expressions and make deductions based on the transformed expression.

Expression $2$

$$ n^2 + 8n + 7 = (n+1)(n+7) = (n+1)({n+2}+5) = (n+1)(\color{red}{n+2}+5) $$ We have isolated the $(n+2)$ term so we substitute again to get:

$$(n+1)(\color{red}{5k}+5) = (n+1)(5(k+1)) = 5 \cdot (n+1)(k+1)$$

Expression $3$

Left as an exercise. Do it to test your understanding.

Expression $4$

$$n^2 - 2n = n(n-2) = n(n+2-4) = n(5k-4) = 5kn - 4n$$

Here, the last expression gives a multiple of $5$ and a remainder of $4n$. Now, based on the given condition, $4n$ is not divisible by $5$ (Why?). Therefore, the expression will always have a remainder when divided by $5$. This implies it is not divisible by $5$.

Answer 4

Using module arithmetic: $$5\mid (n+2) \iff n+2\equiv 0\pmod{5} \Rightarrow \\ n\equiv -2\equiv 3\pmod{5}\\ 2n\equiv 6\equiv 1\pmod{5}\\ 8n\equiv 24\equiv 4\pmod{5}\\ n^2\equiv 9\equiv 4\pmod{5}$$ Hence: $$n^2-4\equiv 4-4\equiv 0\pmod{5} \quad \checkmark\\ n^2+8n+7\equiv 4+4+7\equiv 15\equiv 0\pmod{5} \quad \checkmark\\ n^2-1\equiv 4-1\equiv 3\pmod{5} \quad \times\\ n^2-2n\equiv 4-2\equiv 2\pmod{5} \quad \times$$ For example, take $n=3$: $$n^2-4=5 \checkmark\\ n^2+8n+7=40\checkmark\\ n^2-1=8 \times\\ n^2-2n=3\times$$