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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.

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  • $\begingroup$ but n is an integer not natural number $\endgroup$ – Mary Tom Jan 21 at 14:48
  • $\begingroup$ p would divide n^2-n-2023 irrespective of positive or negative $\endgroup$ – Mary Tom Jan 21 at 14:49
  • $\begingroup$ Is $n$ a positive integer? $\endgroup$ – Mohammad Zuhair Khan Jan 21 at 14:49
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    $\begingroup$ Trial and error seems to work easily enough. You've eliminated $p=2$, so what about $p=3$? Keep going. Since $7\,|\,2023$ you know the answer must be one of $3,5,7$. $\endgroup$ – lulu Jan 21 at 14:51
  • $\begingroup$ is there any way to do it without trial and error $\endgroup$ – Mary Tom Jan 21 at 14:52
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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.

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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 $n$.

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$n^2-n-2023$ 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 $n=1\implies (2n-1)^2= 1\equiv 8093 \mod 7\implies n^2-n-2023\equiv 0\mod 7.$

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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.

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