Find the smallest prime number $p$ such that $p\, | \,n^2-n-2023$ for some integer $n$. Find the smallest prime number $p$ such that $p\, | \,n^2-n-2023$ for some integer $n$.
since $n^2-n =n(n-1)$ is the product of two consecutive integers they must be even so the difference between an even and odd number is always odd so $n^2-n-2023$ is always odd which implies $p$ is not even and the only even prime is $2$ so $p\neq 2$  but after this I do not know what to do please help.
 A: Let $f(n)=n^2-n-2023$ and $f_p(n)=f(n)\pmod{p}$
$\begin{array}{c|cc} 
n & 0 & 1\\  \hline 
f_2(n) & 1 & 1\end{array}$ thus $f_2(n)\neq 0$ and $2$ is not a prime factor of $f(n)$
$\begin{array}{c|ccc} 
n & 0 & 1 & 2\\  \hline 
f_3(n) & 2 & 2 & 1\end{array}$ thus $f_3(n)\neq 0$ and $3$ is not a prime factor of $f(n)$
$\begin{array}{c|ccccc} 
n & 0 & 1 & 2 & 3 & 4\\  \hline 
f_5(n) & 2 & 2 & 4 & 3 & 4\end{array}$ thus $f_5(n)\neq 0$ and $5$ is not a prime factor of $f(n)$
And since $f(1)=-2023=-7\times 17^2$ then $p=7$ is the lowest prime dividing $f(n)$ for some $n$.
A: The answer is 7.
As 2023 is a multiple of 7, it follows that for any $n$ that is a multiple of 7, so
will $n^2-n-2023$ be a multiple of 7. So now it remains to show that $n^2-n-2023$ is not a multiple of 2,3, or 5 for any integer $n$.
However, the prime 5 is not a  multiple for any integer $n$: $2023 \equiv 3 \mod 5$ yet there is no $j \in \mathbb{F}_5$ such that $j^2-j = 3$. This implies that there is no $n$ that satisfies $n^2-n \equiv 3 \mod 5$ [make sure you see why] which implies that there is no $n$ s.t. $n^2-n -2023$ is divisible by 5.
Likewise 3 will not be a multiple for any integer $n$; $2023 \equiv 1 \mod 3$ yet there is no $j \in \mathbb{F}_3 $ such that $j^2-j = 1$. This implies that there is no $n$ s.t. $n^2-n -2023$ is divisible by 3 [make sure you see why].
Meanwhile you can check that $n^2-n-2023$ is always odd for each integer $n$; if $n$ is even it is the sum of two evens and 1 odd, if $n$ is odd it is the sum of 3 odds. This implies that there is no $n$ s.t. $n^2-n -2023$ is divisible by 2.
A: $n^2-n-2023$ is odd, so for prime $p$ we have, modulo $p,$ that  $n^2-n-2023\equiv 0\iff 4n^2-4n-8092\equiv 0\iff (2n-1)^2\equiv 8093.$
Now $8093\equiv 2 \mod 3,$ and $2$ is not  the residue$\mod 3$ of a square, so $p\ne 3.$
And $8093\equiv 3 \mod 5,$ and $3$ is not the residue$\mod 5$ of a square, so $p\ne 5.$
Since $8093\equiv 1\equiv 1^2 \mod 7,$ we have  $n=1\implies (2n-1)^2= 1\equiv 8093 \mod 7\implies  n^2-n-2023\equiv 0\mod 7.$
A: Note that $2023 = 7 \cdot 17^2$
$p=3$
\begin{align}
   n^2-n-2023 \equiv 0 \pmod 3 \\
   n^2+2n-1 \equiv 0 \pmod 3 \\
   (n+1)^2 \equiv 2 \pmod 3
\end{align}
\begin{array}{|c|cc|}
\hline
  n \mod 3 & 0 & 1,2 \\
n^2 \mod 3 & 0 & 1  \\
\hline
\end{array}
The perfect squares modulo $3$ are $0$ and $1$. So there is no solution.
$p=5$
\begin{align}
   n^2-n-2023 &\equiv 0 \pmod 5 \\
   n^2+4n+2    &\equiv 0 \pmod 5 \\
   (n+2)^2+3  &\equiv 0 \pmod 5 \\
   (n+2)^2    &\equiv 2 \pmod 5
\end{align}
\begin{array}{|c|ccc|}
\hline
  n \mod 5 & 0 & 1,4 & 2,3 \\
n^2 \mod 5 & 0 &  1   & 4  \\
\hline
\end{array}
The perfect squares modulo $5$ are $0, 1, 4$. So there is no solution.
$p=7$
\begin{align}
   n^2-n-2023 &\equiv 0 \pmod 7 \\
   n^2-n      &\equiv 0 \pmod 7 \\
   n(n-1)     &\equiv 0 \pmod 7 \\
\end{align}
Clearly $n \equiv 0 \pmod 7$ and $n \equiv 1 \pmod 7$ are solutions.
