[vc_row][vc_column][vc_column_text]
Given the list of the first n primes, 2, 3, 5,…, the product of the primes plus one will be coprime to the list of primes used. For example:
\begin{align*}E_1&=2+1\&=3\end{align*}
and 3 is coprime to 2.
\begin{align*}E_2&=2\times3+1\&=7\end{align*}
and 7 is coprime to 2 and 3.
Will $E_n$ always be a prime number?
Will $E_n$ ever be a square number?[latexpage][/vc_column_text][/vc_column][/vc_row][vc_row][vc_column][lvca_accordion][lvca_panel panel_title="Reveal Solution"]
The first value of $E_n$ that is not a prime number is $E_6$.
\begin{align*}E_6&=2\times3\times5\times7\times11\times13+1\&=30031\&=59\times509\end{align*}
$E_n$ will never be a square number.
$E_n=p_1\times p_2 \times \dots \times p_n +1$
If $E_n$ is a square number this can be written as:
\begin{align*}m^2 &= p_1 \times p_2 \times \dots\times p_n +1\p_1 \times p_2 \times \dots\times p_n &= m^2 -1\&=(m+1)(m-1)\end{align*}
$(m+1)(m-1)$ must either be both odd or both even for this result to hold.
$p_1 \times p_2 \times \dots\times p_n$ has a single even factor of 2. $(m+1)$ and $(m-1)$ cannot both be odd as this would result in $p_1 \times p_2 \times \dots\times p_n$ being odd. $(m+1)$ and $(m-1)$ cannot both be even as this would mean 4 was a factor of $p_1 \times p_2 \times \dots\times p_n$.
Therefore $E_n$ will never be a square number.[/lvca_panel][/lvca_accordion][vc_column_text]
Problem Page Coordinator: Stephen Lee CMath MIMA – Mathematics in Education and Industry
Acknowledgement: The IMA are indebted to MEI for sourcing and supplying Mathematics Today with these well-known puzzles.
First published in Mathematics Today (December 2015)
Given the list of the first n primes, 2, 3, 5,…, the product of the primes plus one will be coprime to the list of primes used. For example:
\begin{align*}E_1&=2+1\&=3\end{align*}
and 3 is coprime to 2.
\begin{align*}E_2&=2\times3+1\&=7\end{align*}
and 7 is coprime to 2 and 3.
Will $E_n$ always be a prime number?
Will $E_n$ ever be a square number?[latexpage][/vc_column_text][/vc_column][/vc_row][vc_row][vc_column][lvca_accordion][lvca_panel panel_title="Reveal Solution"]
The first value of $E_n$ that is not a prime number is $E_6$.
\begin{align*}E_6&=2\times3\times5\times7\times11\times13+1\&=30031\&=59\times509\end{align*}
$E_n$ will never be a square number.
$E_n=p_1\times p_2 \times \dots \times p_n +1$
If $E_n$ is a square number this can be written as:
\begin{align*}m^2 &= p_1 \times p_2 \times \dots\times p_n +1\p_1 \times p_2 \times \dots\times p_n &= m^2 -1\&=(m+1)(m-1)\end{align*}
$(m+1)(m-1)$ must either be both odd or both even for this result to hold.
$p_1 \times p_2 \times \dots\times p_n$ has a single even factor of 2. $(m+1)$ and $(m-1)$ cannot both be odd as this would result in $p_1 \times p_2 \times \dots\times p_n$ being odd. $(m+1)$ and $(m-1)$ cannot both be even as this would mean 4 was a factor of $p_1 \times p_2 \times \dots\times p_n$.
Therefore $E_n$ will never be a square number.[/lvca_panel][/lvca_accordion][vc_column_text]
