Lemma: if $ab=cd$ and $\gcd(a,b)=\gcd(c,d)=1$ and $a,b,c,d \ne 1$ then $a=c,b=d$ or $a=d,b=c$ case 1: $n^2-n+1 \vdots 2 \Rightarrow p(p-1)\vdots 4 \Rightarrow p=2 \Rightarrow n=0$ case 2: $n^2-n+1 \not\vdots 2$ if $n+1 \not\vdots 3$ then $\gcd(2(n+1),n^2-n+1)=\gcd(2(n+1),3n)=1$ and we have $\gcd(p,p-1)=1$ so use the lemma we have $p-1=2(n+1)$ and $p=n^2-n+1$ or $p-1=n^2-n+1$ and $p=2(n+1) \Rightarrow n^2-3n-1=1$ or $-1$ if $n+1\vdots 3$ then $n=3k+2$ put it in the equation and do similarly to case 2

Quote: case 1: $n^2-n+1 \vdots 2 \Rightarrow p(p-1)\vdots 4 \Rightarrow p=2 \Rightarrow n=0$ from $p(p-1)\vdots 4$ it doesn't follow that $p=2$ because it is enough that $p=4k+1$ and by Dirichlet theorem there exists infinitly many such primes

Also, the lemma is wrong: $6\cdot 7=3\cdot 14$

If p=2, n is not a natural number...

Case $1)$ $p\leq n+1$ We get$LHS \leq n^2+n$, so $2(n^3+1) \leq n^2+n$ which is false when $n\in \mathbb{Z}^+$ Case $2)$ $p>n+1>2$, since $n=1$ is not a solution Since $p\mid 2(n+1)(n^2-n+1)$ and $p>n+1$, $p$ is odd number So $p \mid n^2-n+1$ So there exist positive integer $k$ such that $n^2-n+1=pk$ Give us $p=2(n+1)k+1$, so $n^2-n+1=2(n+1)k^2+k$ So $n^2-(2k^2+1)n-(2k^2+k-1)=0$, so $\triangle_n$ is perfect square So $(2k^2+1)^2+4(2k^2+k-1)=4k^4+12k^2+4k-3$ is perfect square But $(2k^2+4)^2=4k^4+16k^2+16 >4k^4+12k^2+4k-3 >(2k^2+3)^2=4k^4+12k^2+9$ when $k>3$ So $k\leq 3$ and we need to check this trivial case

Yes, that's right. If you are interested the answer is $n=20$ and $p=127$

What is the source of this problem ...

Belarusian National Olympiad year 2012

Got n=20 and p=127 for this one

Similar to BMO 2005.

thuanz123 wrote: Lemma: if $ab=cd$ and $\gcd(a,b)=\gcd(c,d)=1$ and $a,b,c,d \ne 1$ then $a=c,b=d$ or $a=d,b=c$ Let, $a=a_1^{\alpha_1}a_2^{\alpha_2}\cdots p_k^{\alpha_k}$ and, $b=b_1^{\beta_1}b_2^{\beta_2}\cdots b_l^{\beta_l}$ where $a_1<a_2<\cdots<a_k$ and $b_1<b_2<\cdots<b_l$ are prime numbers, also, $a_1,a_2,\ldots,a_k$ and $b_1,b_2,\ldots, b_l$ are all positive integers. We mean by $\gcd(a,b)=1$ that, all the $a_i$'s are different to all the $b_j$'s where $i=1,2,\ldots,k$ and, $j=1,2,\ldots,l.$ Now, assume that $\gcd (a,b)=1$, we put for example, $c= b_1^{\beta_1} a_2^{\alpha_2}\cdots p_k^{\alpha_k}$ and, $d=a_1^{\alpha_1}b_2^{\beta_2}\cdots b_l^{\beta_l}.$ Now, you can see that, $ab=cd$ and $\gcd(a,b)=\gcd(c,d)=1$ and $a,b,c,d \neq 1$, but $a ,b\notin \{ c,d\}.$