<s>[INST] Is $ \sqrt{2000!+1}$ a rational number? 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?
 [/INST] We know that $2003$ is prime.
So, by Wilson's theorem, $2002! \equiv -1 \pmod{2003}$
$2002 \equiv -1 \pmod{2003}$.
As such, $2001! \equiv 1 \pmod{2003}$
We can use the Euclidean algorithm to get that $1001 \cdot 2001 \equiv 1 \pmod{2003}$
As such $2000! \equiv 1001 \pmod{2003}$
Which means that $2000! +1 \equiv 1002 \pmod{2003}$.
We want to prove that $1002$ is not a quadratic residue modulo $2003$.
To do that we'll use the law of quadratic reciprocity.
$\left(\frac{1002}{2003}\right) = \left(\frac{2}{2003}\right) \cdot \left(\frac{3}{2003}\right) \cdot \left(\frac{167}{2003}\right)$.
$\left(\frac{2}{2003}\right) = - 1$ because $2003 \equiv 3 \pmod8$
$\left(\frac{3}{2003}\right) = -\left(\frac{2003}{3}\right) = -\left(\frac{-1}{3}\right) = 1$ because both $3$ and $2003$ are congruent to $3$ mod $4$.
$\left(\frac{167}{2003}\right) = - \left(\frac{2003}{167}\right) = -\left(\frac{-1}{167}\right) = 1$ because both $167$ and $2003$ are congruent to $3$ mod $4$.
$\left(\frac{1002}{2003}\right) = -1$ which means $2000! + 1$ is not a perfect square.
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