Q. 124 - Precalculus Enhanced with Graphing Utilities

Question

Pierre de Fermat $\left(1601-1665\right)$ conjectured that the function

$f\left(x\right)=2^{\left(2^x\right)}+1$

for $x=1,2,3,...$ would always have a value equal to a prime number. But Leonhard Euler $\left(1707-1783\right)$ showed that this formula fails for $x=5$. Use a calculator to determine the prime numbers produced by f for $x=1,2,3,4$. Then show that $f\left(5\right)=641\times 6,700,417$, which is not prime.

Step-by-Step Solution

Answer

To show that the formula fails for $x=5$ and also show that $f\left(5\right)=641\times 6,700,417$ is not a prime number first substitute $x=1,2,3,4,5,...$ and solve the equation.

Step 1. Given information.

Consider the given question,

$f\left(x\right)=2^{\left(2^x\right)}+1$

Substitute $x=1$ in the equation,

$$\begin{gathered} f\left(1\right)=2^{2^1}+1 \\ =2^2+1 \\ =4+1 \\ =5 \end{gathered}$$

Substitute $x=2$ in the equation,

$$\begin{gathered} f\left(2\right)=2^{2^2}+1 \\ =2^4+1 \\ =16+1 \\ =17 \end{gathered}$$

Step 2. Find for f 3 , f 4 .

Substitute $x=3$ in the equation,

$$\begin{gathered} f\left(3\right)=2^{2^3}+1 \\ =2^8+1 \\ =256+1 \\ =257 \end{gathered}$$

Substitute $x=4$  in the equation,

$$\begin{gathered} f\left(4\right)=2^{2^4}+1 \\ =2^{16}+1 \\ =65,536+1 \\ =65,537 \end{gathered}$$

Then, the prime numbers produced for f for $x=1,2,3,4,5,17,...,65537$.

Step 3. Find for f 5 .

Substitute $x=5$ in the equation,

$$\begin{gathered} f\left(5\right)=2^{2^5}+1 \\ =2^{32}+1 \\ =4,294,967,296+1 \\ =4,294,967,297 \end{gathered}$$

$f\left(5\right)$ evaluates to $4,294,967,297$ which is not a prime number since it has two factors $641$ and $6,700,417$. That is $4,294,967,297$ is equivalent to $641\times 6,700,417$.

Hence, the formula fails for $x=5$.