Let $n \ge 2$ be integer and $p$ be a prime number such that $n$ and $p$ are non-consecutive integers. Show that if $n \mid (p-1)$ and $p \mid (n^4+n^3-n-1)$, then $4p-3$ is a quadratic number.
Note: An integer $n$ is a called quadratic if there is an integer $x$ such that $n=x^2$.
Attempt:
Since $n \mid (p-1)$, there is a positive integer $b$ with $b \ge 1$ such that $p-1=bn$, i.e., $p=bn+1$. Also, $n \mid (p-1)$ implies that $n \le p-1$. Now, since $p \mid (n^4+n^3-n-1)$ and $n^4+n^3-n-1=(n+1)(n-1)(n^2+n+1)$, then $p \mid (n+1), p \mid (n-1)$, or $p \mid (n^2+n+1)$.
If $p \mid (n-1)$, then $p \le n-1$ so that $$n \le p-1 \le n-2 \implies 0 \le -2,$$ which is impossible.
If $p \mid (n+1)$, then $p \le n+1$ so that $$n \le p-1 \le n \implies n=p-1 \implies p-n=1,$$ which is impossible since $n$ and $p$ are non-consecutive integers. Hence, $p \mid (n^2+n+1)$ which means that there exists a positive integer $c$ with $c \ge 1$ such that $n^2+n+1=cp$. Thus, we obtained that $$n^2+n+1=cp=c(bn+1)=cbn+c. \tag 1$$ By taking modulo $n$ on (1), we have $$c \equiv 1 \pmod n,$$ which means that $c=dn+1$ for some non-negative integer $d$. Substituting this into (1), we have $$n^2+n+1=(dn+1)(bn+1). \tag 2$$ If $d \ge 1$, then $$n^2+n+1= (dn+1)(bn+1) \ge (n+1)^2 = n^2+2n+1 > n^2+n+1,$$ which is impossible. Hence, $d=0$ so that $c=1$ which implies that $$n^2+n+1 = cp = p.$$ Thus, we have $$4p-3 = 4(n^2+n+1)-3 = 4n^2+4n+4-3 = 4n^2+4n+1 = (2n+1)^2,$$ where $2n+1$ is clearly an integer. Therefore, $4p-3$ is a quadratic number, as desired. Q.E.D.
Does the above approach correct? Many thanks in advanced.