Question

Prove that any number of the form \[{{2}^{4n}}-1\]is divisible by 15.

Answer 1

Hint: Here, apply the principle of mathematical induction by taking \[P\left( n \right)={{2}^{4n}}-1\]. First, check for n=1, if it P(1) is divisible by 15 or holds true, then assume that it holds true for n=k and then check if it also holds true for n=k+1, in order to prove that \[{{2}^{4n}}-1\] is divisible by 15

Complete step-by-step answer:
In the question, we have to prove that any number of the form \[{{2}^{4n}}-1\] is divisible by 15.
So here we will apply the concept of principle of mathematical induction. In this method, we will first assume that \[P\left( n \right)={{2}^{4n}}-1\], then we follow the following steps.
Step 1- We will first check that if for n=1, is p(1) divisible by 15, so we have:
\[\begin{align}
  & \Rightarrow P\left( 1 \right)={{2}^{4n}}-1 \\
 & \Rightarrow P\left( 1 \right)={{2}^{4\times 1}}-1 \\
 & \Rightarrow P\left( 1 \right)=16-1 \\
 & \Rightarrow P\left( 1 \right)=15 \\
\end{align}\]
So here this shows that for n=1, the number \[{{2}^{4n}}-1\] is divisible by 15.
Step-2 We will assume that it also holds true for any integer value n=k. So, we have \[P\left( k \right)={{2}^{4k}}-1\]as a multiple of 15, which is further written as:
\[\begin{align}
  & \Rightarrow P\left( k \right)=15x \\
 & \Rightarrow {{2}^{4k}}-1=15x\,\,\,\,\,\,\,\,\,\,(eq-1) \\
\end{align}\]
Here x is the integer number.
Step-3, here will check if \[P\left( k+1 \right)\] is also multiple of 15 or not. So for n=k+1, we have:
\[\begin{align}
  & \Rightarrow P\left( k+1 \right)={{2}^{4\left( k+1 \right)}}-1 \\
 & \Rightarrow P\left( k+1 \right)={{2}^{4k+4}}-1 \\
 & \Rightarrow P\left( k+1 \right)={{2}^{4}}\times {{2}^{4k}}-1\,\,\,\,\,\,\,\,\,\because {{a}^{y+z}}={{a}^{y}}\times {{a}^{z}} \\
 & \Rightarrow P\left( k+1 \right)=16\times {{2}^{4k}}-16+15 \\
 & \Rightarrow P\left( k+1 \right)=16\left( {{2}^{4k}}-1 \right)+15 \\
\end{align}\]
Now here we will use the equation 1 which is \[{{2}^{4k}}-1=15x\,\] and put in the above equation, to get:
\[\begin{align}
  & \Rightarrow P\left( k+1 \right)=16\left( {{2}^{4k}}-1 \right)+15 \\
 & \Rightarrow P\left( k+1 \right)=16\left( 15x \right)+15 \\
 & \Rightarrow P\left( k+1 \right)=15\left( 16x+1 \right) \\
\end{align}\]
Now, here we can see that 16x+1 is also an integer and hence \[15\left( 16x+1 \right)\] is the multiple of 15. So here we have proved that \[P\left( k+1 \right)\] is also the multiple of 15.
Hence, finally using the principle of mathematical induction we can say that any number of the form\[{{2}^{4n}}-1\] is divisible by 15. Hence proved.

Note: It can be noted that n is always an integer and can’t be any fractional number. Also, we can use the method of contradiction in order to prove that \[{{2}^{4n}}-1\] is divisible by 15. Here we will assume that \[{{2}^{4n}}-1\] is not divisible by 15, and then contradicts our assumption by showing that the number\[{{2}^{4n}}-1\] is a multiple of 15, or it is of the form \[{{2}^{4n}}-1=15\,s\], where s is any real number that is integer.