A problem related to divisibility

[Rikudo]

Homework Statement

Let $n, p > 1$ be positive integers and $p$ be prime. Given that $n | p − 1$ and $p | n^3 − 1$,
prove that $4p − 3$ is a perfect square

Relevant Equations

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So, $n\, |\, (p − 1)$ implies $p = nk + 1$ and $p ≥ n + 1$.

Clearly, $p \,|\, n^3 − 1$ implies either :
$p \,|\, n − 1$ (which is impossible, because p cannot be less than $n-1$) or $p \,|\,n^2 + n + 1$.

Now, our main focus is $p\, |\,n^2 + n + 1$.
Since $p = nk + 1$, this will become $(nk+1) \,|\,n^2 + n + 1$

In order to solve the problem, I think I must find $p$ in terms of $n$. For this reason, I need to find the possible values of $k$ first.
So:

Because the remainder of the division of $(n^2 + n + 1)$ by $(nk+1)$ is $1-(\frac {k-1}{k^2})$ and must be 0, It can be concluded that
$$1-(\frac {k-1}{k^2})=0$$
$$k^2-k+1 = 0$$
Strangely, after solving the above equation,the value of $k$ will be imaginary number.
What is wrong here?

 

[Office_Shredder]

I think the problem is you found the remainder doing division in the ring of rational polynomials over n, but actually care about division in the integers. Those aren't going to be the same thing.

Edit to add: thinking about this a bit more, they are closer than I realized, but definitely not the same.

If you happen to factor $p(n)=q(n)k(n)+r(n)$ when dividing $p(n)$ by $q(n)$, if $k(n)$ and $r(n)$ are guaranteed to be integers, and $r(n)<q(n)$, then you do have a good integer factorization. This happens for lots of natural examples, for example $p=n^2$ and $q=n+1$ gives $k=n-1$, $r=1$ and when you divide $n^2$ by $n+1$ in the integers you do in fact get a remainder of 1.

In your scale, $k(n)$ is almost certainly not an integer. It's closer to something like divide $n^2$ by $5n$. Then you get $k=\frac{1}{5}n$, $r=0$ but the remainder has nothing to do with integer division unless $k$ happens to be an integer.
Or another example, divide $n^2$ by $n+100$. This gives $k=n-100$, $r=10000$. But when you pick $n=30$ and divide $900$ by $103$, the remainder isn't $10000$, because we failed the requirement that $r<q$

 

[Rikudo]

So, to find $k$, should I make use of the quotient instead of the remainder?

The quotient will be : $\frac n k + \frac {k-1} {k²}$Here is what I thought initially:
At first glance, the only possible value of $k$ is 1. Otherwise, the second term will become a fraction (consequently, the quotient will not be an integer)However, now that I think about it, isn't it possible for the quotient to be an integer even though if both term is a fraction?

 

[PeroK]

I would have checked this out for the first few primes to see whether I could spot a pattern for $p, n$ and $4p -3$.

It also helps to convince yourself that what you are trying to prove might actually be true!