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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?

Some Guy
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    According to http://oeis.org/A146968 the only values of $n$ with $n < 10^9$ such that $n!+1$ is a perfect square are $n=4,5,7$. I think it's reasonable to expect that these are the only such values, but that seems to be open. – Nate Apr 02 '21 at 16:43
  • What does probability have to do with anything here? as in "I realised that probably it is not a rational number" – David G. Stork Apr 02 '21 at 16:50
  • @DavidG.Stork What probability? – Some Guy Apr 02 '21 at 16:50
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    There's also a Wikipedia article, quoting recent results that improve the bound to $10^{15}$. – Troposphere Apr 02 '21 at 16:50
  • Just a remark, a rational number is $\frac{n}{m}$. Hence, the proof by contracdition should begin with $$2000!+1 = \frac{n^2}{m^2}$$ – NN2 Apr 02 '21 at 16:52
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    @NN2: Why? It's clearly an integer, so $m=1$ and can therefore be omitted. – Asaf Karagila Apr 02 '21 at 16:53
  • @DavidG.Stork it was a manner of speaking. It simply didn`t look like an usual square. –  Apr 02 '21 at 16:54
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    This is a purely computational problem, and can be solved purely computationally. The only integers with rational square roots are squares. Then to determine this, it suffices to compute the square root of $2000! + 1$ with enough precision (it seems that 3000 digits is enough, as I just did it on my machine). Or alternately, find the two bounding square numbers that surround $2000! + 1$, which can be done using only big integer arithmetic and a bisection-type algorithm. – davidlowryduda Apr 02 '21 at 16:55
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    @Troposphere Indeed. Although the advantage of OEIS over Wikipedia is that I arrived at that page by just knowing that 4!+1 and 5!+1 work. – Nate Apr 02 '21 at 16:55
  • @AsafKaragila A rational number is $\in \Bbb Q$ and is equal to $\frac{n}{m}$ – NN2 Apr 02 '21 at 16:55
  • @davidlowryduda as this might sound primitive and possibly silly, the problem poser requires a mathematical proof rather than a computer-assisted proof. –  Apr 02 '21 at 16:56
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    @NN2: Yes, I know what is a rational number. But a rational number is the square of an integer if and only if it is an integer. – Asaf Karagila Apr 02 '21 at 16:56
  • @AsafKaragila Ok, you can prove it, but this result $\sqrt{n} =\frac{a}{b} \implies b = 1 $ is not something evident. – NN2 Apr 02 '21 at 16:59
  • @Nate: True -- and I did find the Wikipedia article because OEIS told me the name of the problem. It's a case of synergy between the sources, not a competition :-) – Troposphere Apr 02 '21 at 17:03

4 Answers4

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Here is a proof that can be done by hand, though certain steps are simplified by having a computer.

  1. First, note that $2003$ is prime. (You could check this by noting that no primes below $50$ divide it).
  2. 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 $$2000! \equiv 1001 \bmod 2003.$$
  3. 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 $$ \left( \frac{1002}{2003} \right) = \left( \frac{2}{2003} \right) \left( \frac{3}{2003}\right) \left( \frac{167}{2003} \right). \tag{1}$$
  4. As $2003 \equiv 3 \bmod 8$, we know that $2$ is not a square mod $2003$. This is the first symbol.
  5. 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 $2003$.
  6. 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 $2003$.
  7. 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.

davidlowryduda
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    You can save a lot of work by noting that $1002 = 2^{-1} \pmod{2003}$, and therefore 1002 is a square if and only if 2 is. A similar idea simplifies step 2 $$2000! \cong -1 \cdot (-1)^{-1} \cdot (-2)^{-1} \cong 2002 \cdot -1 \cdot -1/2 \cong 1001 \pmod{2003}.$$ – Ravi Fernando Apr 02 '21 at 17:45
  • @RaviFernando That is an excellent observation. – davidlowryduda Apr 02 '21 at 17:56
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    @davidlowryduda Thank you for this beautiful answer!!! I accepted it! –  Apr 06 '21 at 06:32
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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.

Oussema
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    @Quessema That's the idea I had too (I just didn't have time to finish the quadratic reciprocity law computations), another way to find that $1001 \cdot 2001 = 1 (\mod 2003)$ is to note that $2 \cdot 1002 = 1 (\mod 2003)$ and $2001=-2 \rightarrow 1/2001=-1/2 = -1002=1001$. – Omer Apr 02 '21 at 17:12
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    Ah, another with the same proof. It turns out this takes me at least 5 minutes to write out. – davidlowryduda Apr 02 '21 at 17:12
  • @davidlowryduda You did take more time doing the careful formatting and elaborating the steps. +1 from me. – Oussema Apr 02 '21 at 17:14
  • Some (...) deleted my answer, a vote to undelete would be appreciated. – Igor Rivin Apr 02 '21 at 18:32
  • @IgorRivin sorry I don't have enough reputation yet – Oussema Apr 02 '21 at 18:35
  • No problem, thank you! – Igor Rivin Apr 02 '21 at 18:38
  • @Oussema this is ok so I upvoted it. You wrote a good answer. Thank you! –  Apr 06 '21 at 06:33
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I hope this doesn’t get downvoted but I’ll write it down nonetheless: it can be computed (wolfram did it) that $2000!+1$ is congruent to $371$ mod $2017$ and $2017$ is a prime. But $371^{1008}$ is $-1$ mod $2017$ (again, by wolfram), so that $371$ is not a square mod $2017$, and thus $2000!+1$ cannot be a square (and rational square roots of integers are integers, which completes the proof).

Aphelli
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In Mathematica,

IntegerQ[Sqrt[2000!+1]] returns False, so you are good.

ADDENDUM For those who think there should be a cute "human" proof, read https://www.wikiwand.com/en/Brocard%27s_problem

Igor Rivin
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    I'm not the downvoter. However, Mathematica can be wrong; I have already seen such cases. Moreover, this is not a proof. – mathcounterexamples.net Apr 02 '21 at 16:43
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    @Ooussema That IS a proof. As for "mathematical can be wrong" - true, but not for arithmetic. – Igor Rivin Apr 02 '21 at 16:44
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    @IgorRivin Not a very insightful proof, however – Some Guy Apr 02 '21 at 16:46
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    @IgorRivin Hi Igor. I didn't downvote your answer. However, it is not a proof. It is rather an observation. A proof would be appreciated! Thank you. –  Apr 02 '21 at 16:51
  • @AnatolDenicula It IS a proof. Just as to prove 17 is not a square you take its square root, so for $2000!+1.$ If you had the time, you could do it yourself (it does not take THAT long. – Igor Rivin Apr 02 '21 at 16:53
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    Also, it's not difficult to write a program for oneself that does enough bignum arithmetic from scratch to verify the claim that $a^2 < 2000!+1 < (a+1)^2$ for some concretely given $a$. It doesn't have to be particularly efficient bignum arithmetic for these sizes -- so just use base 10 -- and finding an $a$ to input into the program can be done with standard software, which we don't need to trust as long as it happens to produce the $a$ that works. – Troposphere Apr 02 '21 at 17:09
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    If what one seeks is just the answer. Then I absolutely agree. But sometimes the point is to learn how to tackle a problem. – Oussema Apr 02 '21 at 17:21
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    I don't quite understand why this answer was deleted. I've undeleted this answer. I've also cleaned some of the comments on this answer. I should also note that the OP did not specify that they wanted a general proof, but instead only asked about the particular case of $\sqrt{2000! + 1}$. This is easily answerable by direct computation, and this answer is one particular direct computation. – davidlowryduda Apr 05 '21 at 17:15
  • @IgorRivin Your answer should not be deleted. It is a good observation and a solution using the computer. It is valid and I upvoted it. However, I accepted another one, a "paper and pencil" one, but your answer is good too and it should stay. It is an alternative and a note. Thank you very much! –  Apr 06 '21 at 06:35