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.
 A: First we rewrite the congruence:
$$4x(x+1)=4\pmod{4n}\quad\Leftrightarrow\quad (2x+1)^2=5\pmod{4n}\ .$$
Now write $y=2x+1$ and consider various cases.


*

*If $n$ is even then the above implies $y^2=5\pmod8$, which has no solution.

*If $5^2\mid n$ then we get
$$\eqalign{y^2=5\pmod{25}\quad
  &\Rightarrow\quad 5\mid y^2\cr
  &\Rightarrow\quad 25\mid y^2\cr
  &\Rightarrow\quad 25\mid 5\cr}$$
and again there is no solution.

*So for solutions to exist, $n$ is a product of $5$ (possibly) and prime powers $p^\alpha$ where $p$ is not $2$ or $5$.  There is a solution iff
$$y^2=5\pmod{p^\alpha}$$
has a solution for every such prime power, which can be proved to be equivalent to
$$y^2=5\pmod p$$
having a solution for every $p\mid n$ (except $p=2,5$).  Using the Legendre symbol one can show that this comes down to the following.

The congruence $x(x+1)=1\pmod n$ has a solution if and only if $n$ is a product of primes in which $5$ occurs only once (or not at all), and every other prime is congruent to $1$ or $4$ modulo $5$.


Here is a table of some low values of $n$.  I have listed "ok" if the congruence has a solution, otherwise I have given a "bad" prime factor of $n$.
$$\def\ok{{\rm ok}}
  \matrix{3&5&7&9&11&13&15&17&19&21&23&25&27&29&31&\cdots&55\cr
          3&\ok&7&3&\ok&13&3&17&\ok&3&23&5&3&\ok&\ok&\cdots&\ok\cr}$$
Note that $p=5$ is "bad" for $n=25$ because it occurs twice, but it is "ok" for $n=5$ and $n=55$ because it only occurs once.
A: 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 $n$.
A: As  lab bhattacharjee wrote
(working $\bmod n$),
If $m(m+1) = 1$
then
$4m^2+4m=4$
or
$5
=4m^2+4m+1
=(2m+1)^2
$.
Therefore,
if $5$
is a quadratic residue mod $n$,
there is (at least)
one solution.
According to
https://en.wikipedia.org/wiki/Quadratic_residue,
for a prime $p$,
$5$ is a quadratic residue
mod $p$
if and only if
$p \equiv
1, 4 (\bmod 5)
$.
For composite $n$,
see
https://en.wikipedia.org/wiki/Jacobi_symbol.
The table there
titled
"Table of values"
shows which
$k$ have
$5$ as a quadratic residue.
A: Clearly, the condition $x(x+1)\equiv1\pmod{n}$ cannot hold if $n$ is even (because $x(x+1)$ is itself even).
Now suppose $n$ is odd.
The OP can be reformulated as follows : find the integers $n\ge2$ such that the equation $x^2+x-\bar{1}=\bar{0}$ has at least a solution in the ring $\mathbb{Z}/n\mathbb{Z}$.
(We denote by $\bar a$ the congruence class of $a$ mod. $n$)
Note that $\bar2$ is invertible in that ring (because $gcd(2,n)=1$).
The discriminant of this quadratic equation is $\Delta=\bar5$ and the question becomes : find the integers $n\ge2$ such that $\bar5$ is a square.
The answers is given by the Legendre symbol. For example, si $n$ is an odd prime and $n\neq5$, then :
$$\left(\dfrac{5}{n}\right)=(-1)^{\left\lfloor\frac{n+2}5\right\rfloor}$$
In the general case, we can use de quadratic reciprocity law to compute $\left(\dfrac{5}{n}\right)$ and decide it is $1* or not
