Answer
Consider another formula to find the prime numbers,
\[f\left( n \right)={{n}^{2}}-79+1601\]
Now, verify the formula for prime numbers by substituting\[\left( n=0,1,2,3,4,5.....79 \right)\].Then, the prime numbers are shown below in the table,
Work Step by Step
Now, check the above formula for\[n=80\].
\[\begin{align}
& f\left( 80 \right)={{80}^{2}}-79\times 80+1601 \\
& =1681 \\
& ={{41}^{2}}
\end{align}\]
The prime number is the number that divisible by itself or 1. From the above table, the above formula shows the prime numbers between \[\left( n=0 \right)\]to\[\left( n=79 \right)\]. But at \[\left( n=80 \right)\] the formula fails.
Hence, from the above observation, no unique formula is true for all prime numbers.
