Question

Prove that 4 is a factor of$7^n-3^n$ for every positive integer n.

Short answer

It is proved that4 is a factor of$7^n-3^n$ for every positive integern.

Step-by-step solution

Consider the algebraic formula as follows

It is known the algebraic formula for factoring $a^n-b^n$is given by

$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+a^{n-3}{b}^2+.....+a{b}^{n-2}+b^{n-1}\right)$

Check the expression for n = 1

Consider that n be any positive integer and$p\left(n\right)=7^n-3^n$ .

Substitute1 forn in the equation$p\left(n\right)=7^n-3^n$ as follows:

$\begin{array}{rcl}p\left(1\right)& =& 7^1-3^1\\ & =& 7-3\\ & =& 4\end{array}$

Hence,4 is divisible by itself; this implies that at$n=1$ , it is true.

Consider the expression for n = m

Now, consider it is true for$n=m$ . Then$p\left(m\right)=7^m-3^m$ is divisible by4, that is,$p\left(m\right)\overline{)4}$ . So,

$p\left(m\right)=4k$

.

Here,k is an integer.

Check the expression for n = m + 1.

Consider $n=m+1$. Then$p\left(m+1\right)=7^{m+1}-3^{m+1}$ .

Now, apply the algebraic formula to factor $p\left(m+1\right)=7^{m+1}-3^{m+1}$as follows:

$\begin{array}{rcl}p\left(m+1\right)& =& \left(7-3\right)\left(7^{\left(m+1\right)-1}+7^{\left(m+1\right)-2}·3+7^{\left(m+1\right)-3}·3^2+....+7·3^{\left(m+1\right)-2}+3^{\left(m+1\right)-1}\right)\\ & =& 4\left(7^m+7^{m-1}·3+7^{m-2}·3^2+....+7·3^{m-1}+3^m\right)\end{array}$

This implies that$p\left(m+1\right)\overline{)4}$ .

Hence, for any positive integer n,$7^n-3^n$ is divisible by4.