TITLE: Is $ \sqrt{2000!+1}$ a rational number?
QUESTION [13 upvotes]: Is $
\sqrt{2000!+1}$  a rational number?
This may seem trivial, but as I wrote $2000!+1=n^2$ for $n\in\mathbb{N}$, I realised that it probably is not a rational number and that I cannot build a constructive proof, because $n^2-1>2^{2000}$ as from here Prove by induction that $n!>2^n$
and also, as $n!<(\frac{n+1}{2})^{n}$ $n! \leq \left( \frac{n+1}{2} \right)^n$ via induction and  $n^2-1<(1001+\frac{1}{2})^{2002}$ and these are already extremely hard tot tackle. Any help, please?

REPLY [28 votes]: Here is a proof that can be done by hand, though certain steps are simplified by having a computer.

First, note that $2003$ is prime. (You could check this by noting that no primes below $50$ divide it).
It follows from Wilson's Theorem that $2002! \equiv -1 \bmod 2003$. We then also have that
$$ 2000! \equiv -1 \cdot (2002)^{-1} (2001)^{-1} \bmod 2003, $$
and $2002^{-1} \equiv 2002 \bmod 2003$ (as this is just $-1$). A bit more work, perhaps using the extended Euclidean algorithm, shows that $2001^{-1} \equiv 1001 \bmod 2003$. Thus
$$2000! \equiv 1001 \bmod 2003.$$
We thus have that $$ 2000! + 1 \equiv 1002 \bmod 2003.$$ If we could show that $1002$ is not a square mod $2003$ (it's not), then we'll be done. To do this, we can use quadratic reciprocity. Namely we consider the Legendre symbol
$$ \left( \frac{1002}{2003} \right) = \left( \frac{2}{2003} \right) \left( \frac{3}{2003}\right) \left( \frac{167}{2003} \right). \tag{1}$$
As $2003 \equiv 3 \bmod 8$, we know that $2$ is not a square mod $2003$. This is the first symbol.
For $3$, we use quadratic reprocity. The sequence of steps goes
$$
\left( \frac{3}{2003} \right) = -\left( \frac{2003}{3} \right) = - \left( \frac{2}{3} \right) = 1.
$$
Thus $3$ is a square mod $2003$.
For the last one, the sequence of steps goes
$$
\left( \frac{167}{2003} \right) = - \left( \frac{2003}{167} \right)
= -\left( \frac{166}{167} \right),
$$
which we should recognize as asking if $-1$ is a square mod $167$. As $167 \equiv 3 \bmod 4$, it's not a square. Thus $167$ is a square mod $2003$.
We can now conclude. The line in $(1)$ evaluates to
$$ -1 \cdot 1 \cdot 1 = -1,$$
and thus $2000! + 1$ is not a square mod $2003$. And thus it's not a square.


Ravi Fernando pointed out an observation that gives an enormous simplification in the comments. The observation is that $1002 \cdot 2 \equiv 2004 \equiv 1 \bmod 2003$, and thus $1002 = 2^{-1} \bmod 2003$. Thus $1002$ is a square if and only if $2$ is a square (mod $2003$). As $2003 \equiv 3 \bmod 8$, $2$ is not a square, and thus $1002$ is not a square.