How to prove the number is a prime? A natural number $n$ has the property that if $d$ divides $n$ then $d+1$ divides $n+1$. Show that $n$ must be a prime.
 A: Let $n=ab$ with $a\ge b\gt0$.  If $(a+1)\mid(n+1)$, then $ab+1=(a+1)r$ for some $r\gt0$, which implies $r=1+ah$ for some $h\ge0$.  But this leads to
$$b=ah+h+1$$
and the assumption $a\ge b$ implies $h=0$, so that $b=1$.  Thus $n$ must be prime.
A: Assume $n > 2$.  If $n = 1$ or $n= 2$ the result is clear.
Let $n = pq$.  Wolog $p \le q$.
Assume $q+1|n+1$.  Then as $0< q+1 \le n+1$, there exist an integer $k \ge 1$ such that $k(q+1) = kq + k = n+1 = pq + 1$.
So $k = q(p - k) + 1 \ge  $ so $p \ge k$.
$k(q + 1) = (q(p-k) + 1)(q+1) = q^2(p-k) + q  + q(p-k) + 1 = n+1$ so
$q[q(p-k) + 1 + (p-k)] = n = pq$
$q(p-k) + (p-k) + 1 = p$
If $p > k$ then $p = q(p-k) + (p-k) + 1 > q + 2$ so $p > q$ which contradicts our assumption.
So $p \le k$ but we know $p \ge k$ so $ p = k$.
So $p = q(p-k) + (p-k) + 1 = 1$.  So $p = k = 1$ and $q = n$.  
Thus if $n$ has the property then the only possible factors of $n$  are $1$, $n$.
A: Assuming $\exists$ such a $d$ which divides $n$ and $\neq 1$ or $n$.
Also,
$$\text{if } d+1|n+1 \Longleftrightarrow \gcd(d+1, n+1) \neq 1$$
But, $$\gcd(d+1,n+1)=\gcd(d+1-n-1, n+1) = \gcd(d-n, n+1) = \gcd(d(1-k), dk+1)$$
Hence, $\gcd(d+1,n+1)= 1$ which is a contradiction.
So we can claim that no such $d$ exists and hence $n$ is a prime.
Hope that helped.
