Is the product of two consecutive integers $1$ $\pmod n$? Is there a simple test for $n$ to determine if there exists an integer $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. For example, $n$ $=$ $3$ and $n$ $=$ $7$, there are no integers $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. But for $n$ $=$ $5$ and $n$ $=$ $11$, there are integers $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. There are namely,
$x = 2$, 
$2*3$ $=$ $1$ $\pmod 5$
$x = 7$, 
$7*8$ $=$ $1$ $\pmod {11}$
It is confusing to find out which integers $n$ have this property. Maybe there is a special (mathematical) property these numbers have. Help is appreciated. Thanks.

You are asking when the polynomial $x^2+x-1$ has a root mod $n$.  Since $x(x+1)$ is always even, clearly $n$ must be odd for such an $x$ to exist.  In that case $2$ is invertible mod $n$, and so by the quadratic formula $x^2+x-1$ has a root mod $n$ iff the discriminant $5$ has a square root mod $n$.
So you are asking for what odd integers $n$ is $5$ a square mod $n$.  By the Chinese remainder theorem, $5$ is a square mod $n$ iff it is a square mod $p^k$ for each prime power $p^k$ appearing in the prime factorization of $n$.  For $p\neq 2,5$, $n$ is a square mod $p^k$ iff $5$ is a square mod $p$ (for instance, by Hensel's lemma).  By quadratic reciprocity, $5$ is a square mod $p$ for odd $p$ iff $p$ is a square mod $5$, i.e. iff $p$ is $0,1,$ or $4$ mod $5$. The case $p=2$ does not matter since $n$ must be odd, and $5$ is a square mod $5^k$ iff $k\leq 1$.
Putting it all together, we find that $x^2+x-1$ has a root mod $n$ iff every prime factor of $n$ is $0,1,$ or $4$ mod $5$ and $25$ does not divide $n$.
