Question

Prove that ${{3}^{2n}}+7$ is multiple of 8 where $n\in N$.\[\]

Answer 1

Hint: We use the method of induction to prove the given statement${{P}_{n}}:8|{{3}^{2n}}+7$. We prove ${{P}_{n}}$ is true for $n=1$ that is ${{P}_{1}}$ is true, assume that ${{P}_{n}}$ is true for $n=k$ that is ${{P}_{k}}$ is true and prove that ${{P}_{n}}$ is true for $n=k+1$ that is ${{P}_{k+1}}$ is true.\[\]

Complete step-by-step solution:
We know from the induction hypothesis that a statement defined on any natural number $n$ as ${{P}_{n}}$ is true for all values of $n$ if\[\]
1. The statement is true for initial or base value for ex $n=0,1$ or ${{P}_{1}}$ is true.\[\]
2. The statement if assumed to be true for $n=k$(${{P}_{k}}$ is true) for some natural number $k\le n$ will induct the consecutive statement to be true which means the statement is true for $n=k+1$(${{P}_{k+1}}$ is true)\[\]
So we have to prove ${{P}_{1}}$ is true, assume ${{P}_{k}}$ is true and using ${{P}_{k}}$ have to show ${{P}_{k+1}}$ is true. So let us denote the given statement as${{P}_{n}}$. So we have${{P}_{n}}:$ ${{3}^{2n}}+7$ is multiple of 8 or
\[{{P}_{n}}:8|{{3}^{2n}}+7\]
Now let's check if the statement is true for the initial value $n=1$ which if ${{P}_{1}}$ is true or not. We have
\[{{3}^{2\left( 1 \right)}}+7={{3}^{2}}+7=16\]
We see that 16 is divisible by 8 which means ${{P}_{1}}$ is true. Mathematically, it is true that
\[{{P}_{1}}:8|{{3}^{2\left( 1 \right)}}+7\]
Now let us assume that the statement is true for $n=k,k\le n$ which means it is divisible by 8 or
\[{{P}_{k}}:8|{{3}^{2k}}+7\]
Now consider for integer $p$ and have,
\[\begin{align}
  & {{3}^{2k}}+7=8p \\
 & \Rightarrow {{3}^{2k}}=8p-7.....\left( 1 \right) \\
\end{align}\]
Now we have to prove ${{P}_{k+1}}$ is true which means ${{3}^{2\left( k+1 \right)}}+7$ is divisible by 8 which means ${{3}^{2\left( k+1 \right)}}+7=8q$ for some integer $q$. We have
\[\begin{align}
  & {{3}^{2\left( k+1 \right)}}+7={{3}^{2k+2}}+7 \\
 & ={{3}^{2}}\cdot {{3}^{2k}}+7 \\
\end{align}\]
Lets us substitute the value of ${{3}^{2k}}$ obtained in equation (1) and have,
\[\begin{align}
  & =9\left( 8p-7 \right)+7 \\
 & =72p-63+7 \\
 & =72p-56 \\
 & =8\left( 9p-7 \right) \\
\end{align}\]
Let us choose $9p-7=q$ and have
 \[\therefore {{3}^{2\left( k+1 \right)}}+7=8q\]
Hence ${{P}_{k+1}}$ is true which means the stamen is true for all $n\in N$.\[\]

Note: We could choose $9p-7=q$ because by closure law subtraction is closed within the set of integers. We can alternatively prove using binomial expression taking ${{3}^{2n}}+7={{\left( 8+1 \right)}^{n}}+7$ and expanding with $x=8,y=1$ as ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+...+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$ . We simplify to express ${{3}^{2n}}+7$ as a multiple of 8.