6th grade question

Question

I got this problem from some 6-th grade kid:

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If $n=(a+b)(b+c)(c+a)$, where $a,b,c \in \mathbb{N}^*$, then prove that

$$\frac{2^{n+1}+7^{n-1}}{15} \in \mathbb{N}$$

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It has been some time since I tackled problems like this one so I am a bit rusty on the subject.

I tried to first see if expanding the expression of $n$ would give rise to something that can related to the expression that has to be proven. I'll put it below although it is just basic algebra.

$$n = 2abc + a^2(b+c)+ b^2(a+c) + c^2(a+b)$$

Having no hint from that result I went to the expression that has to be proven. I wrote it as

$$14\cdot 2^n + 7^n = k\cdot 105$$

where $k\in \mathbb{N}^*$. One hint came from this, but it did not deliver a solution.

I noticed that $15=2^3 + 7$ from which I can write $15 \cdot k = k(2^3 + 7)$. From this, for $k = 1$ and $k=7$, I could check by hand if I can find a value for $n$ such that it satisfies the equation that has to be proven. I got the values $n=1$ and $n=2$. However, those values cannot be written in as it is required by the first relation.

Next I thought that putting numbers for the values $a,b,c$ could give some clues. Unfortunately brute force with a calculator fails since the smallest value for $n$ is $8$, and the next one is $18$.

Note 1: This last step was also meant to check if by pure luck I can find a value for $n$ where the equation does not hold, thus solving the problem like a boss by mentioning "The problem is incorrect or it has a typo".

Note 2: I am interested in a solution that is fit for a 6-th grader, however any other level (e.g. undergraduate) should be good (I'm not a mathematician, but an engineer).

Answer 1

This answer is perfectly suitable for a $6$-grader.

First of all, note that $n$ is always even. To accept it, note that $n$ can be odd if and only if each one of $a+b$, $b+c$ and $c+a$ must be odd. If any of $a$, $b$ and $c$ is odd, then the other two must be even, implying their sum is even. Hence, $n$ is definitely even. If this sounds confusing, remember

$$\begin{align} \mathrm{odd}\times \mathrm{odd}&=\mathrm{odd}\\ \mathrm{odd}\times \mathrm{even}&=\mathrm{even}\\ \mathrm{even}\times \mathrm{even}&=\mathrm{even}\end{align}$$

and

$$\begin{align} \mathrm{odd}+\mathrm{even}&=\mathrm{odd}\\ \mathrm{odd}+\mathrm{odd}&=\mathrm{even}\\ \mathrm{even}+\mathrm{even}&=\mathrm{even}\end{align}$$

Now, comes the second part of the question. Note that $$\frac{2^{n+1}+7^{n-1}}{15}\in\mathbb{N}\equiv 15|(2^{n+1}+7^{n-1})$$ (If you don't know the statement means that $2^{n+1}+7^{n-1}$ is exactly divisible by $15$.)

Assume, for simplicity, $x=2^{n+1}+7^{n-1}$.

To prove this, we shall do this in two parts - a number is divisible by $15$ if and only if it is divisible by both $3$ and $5$.

A number is divisible by $5$ if and only if it ends in $5$ or $0$.

Thus, $x$ is always divisible by $5$.

Now, let's observe the remainder left by $2^{n+1}$ when divided by $3$.

Thus, $2^{n+1}$ always leaves a remainder of $2$ when divided by $3$.

Now, consider $7^{n-1}$.

Together, $x$ will make a remainder of $3$, which when divided by $3$ gives remainder $0$.

Thus, $x$ is exactly divisible by $3$.

Altogether, we can say that $x$ is divisible by $15$, as required.

Hope this helps :)