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I really need some help at this problem:
Let $p$ be a prime number and $n$ a natural number, $n\ge2$ such that $n \mid p-1$ and $p \mid n^3-1$. Prove that $4p-3$ is a square.
So $p \mid (n-1)(n^2+n+1)$
What if $p \mid n-1$? Treating the cases wasn't too efficient. I was thinking about Fermat's theorem but it didn't helped really much.
A hint would be really apreciated. Thanks!
Note $p \mid n-1$ is impossible because $n \le p-1$, so we have $p \mid n^2+n+1$.
Since $n \mid p-1$, we can write $p = an+1$ for some integer $a \ge 1$. Since $p \mid n^2+n+1$, we can write $$n^2 + n + 1 = bp = b(an+1)$$ for some integer $b \ge 1$.
Reducing modulo $n$ gives $1 \equiv b \pmod{n}$, so write $b = rn+1$ for some integer $r \ge 0$. Putting this in the above equation gives $$n^2 + n + 1 = (rn+1)(an+1).$$
If $r \ge 1$, then $(rn+1)(an+1) \ge (n+1)^2 > n^2+n+1$, which is a contradiction. So $r = 0$, $b = 1$, and $n^2 + n + 1 = p$, so we get $4p-3 = (2n+1)^2$.
Hint: If $p\mid n-1$ and $n\mid p-1$, then we also have $p\leq n-1$ and $n\leq p-1$.
Given that $p\mid(n^3-1)$, thus we have $p\mid(n-1)$ or $p\mid(n^2+n+1)$. We also have $n\mid(p-1)$. Let us make cases.
Case 1: $p\mid(n-1)$ and $n\mid(p-1)$. Since $p-1\ge 1$ and $n>1$. Thus we can conclude that $p\mid(n-1)\implies p\le n-1$ and $n\mid(p-1)\implies n\le p-1$. Thus we have $n\le p-1\le n-2\implies n\le n-2\implies 0\le-2$, which is a clear contradiction. Hence $p\not\mid(n-1)$.
Case 2: $p\mid(n^2+n+1)$ and $n\mid(p-1)$. Thus, $\exists k\in\mathbb{Z},$ such that $pk=n^2+n+1$ and $\exists q'\in\mathbb{Z}$, such that $p=nq'+1$. Now $n^2+n+1\equiv 1\pmod n$ and $p\equiv 1\pmod n\implies pk\equiv k\pmod n\implies k\equiv pk=n^2+n+1\equiv 1\pmod n\implies k\equiv 1\pmod n.$
Thus $\exists q\in\mathbb{Z}$, such that $k=nq+1$. Hence we have $$p=nq'+1 \text{ and } k=nq+1\\ \implies pk=n^2qq'+n(q+q')+1=n^2+n+1\\\implies n(1-qq')=q+q'-1.$$
Now $p=nq'+1\implies nq'=p-1$ and $p-1\ge 1$. Thus $nq'\ge 1$. Now since $n>1$, hence $q'\ge 1$. This again implies that $nq'>1\implies p>2\implies p\ge 3.$
Now $pk=n^2+n+1>3\implies pk>3.$ Now $p\ge 3\implies k\ge 1.$ Now again we have $k=nq+1$. Since, $k\ge 1\implies nq+1\ge 1\implies nq\ge 0\implies q\ge 0$ $(\because n>1).$
Thus $q\ge 0,q'\ge 1\implies qq'\ge 0.$ Also $q+q'\ge 1\implies q+q'-1\ge 0.$ This implies that $n(1-qq')=q+q'-1\ge 0\implies n(1-qq')\ge 0\implies 1-qq'\ge 0 \implies qq'\le 1.$
Therefore, we simultaneously have $qq'\ge 0$ and $qq'\le 1$, which implies that $qq'=0$ or $qq'=1$.
Let us make cases.
Case 1: $qq'=1\implies q=1$ and $q'=1$. Thus $q+q'-1=1$ and $n(1-qq')=0\implies n(1-qq')\neq q+q'-1$, which is a contradiction to the fact that $n(1-qq')=q+q'-1$. Hence $qq'\neq 1$.
Case 2: $qq'=0\implies q=0$. Thus $n=q'-1\implies q'=n+1$, which in turn implies that $p=n(n+1)+1.$ Thus $4p-3=4n(n+1)+1=4n^2+4n+1=(2n+1)^2$.
Hence after analyzing all the cases we can conclude that $4p-3$ is a perfect square and is equal to $(2n+1)^2$, and we are done.
