# Prove: $\lfloor \sqrt{n!}\rfloor\nmid n!$ [closed]

$n$ is an integer greater than $7$. How does one go about proving that $\lfloor \sqrt{n!}\rfloor\nmid n!$.

## closed as off-topic by TheGeekGreek, Daniel W. Farlow, Ross Millikan, heropup, user26857Feb 8 '17 at 23:21

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• Relevant: math.stackexchange.com/q/6369/73324 – vadim123 Feb 8 '17 at 18:12
• the thing to do is calculate this for $n$ up to something modest such as 20, including factoring of both quantities. The catch is that $n!$ is divisible only by primes no larger than $n,$ and the exponents are entirely predictable. – Will Jagy Feb 8 '17 at 18:15
• – i9Fn Feb 8 '17 at 18:26
• @Maazul You have it backwards, your problem is at least as strong as Brocard. Suppose Brocard is false, then there is an $n>7$ and $k$ such that $n!=k^2-1$. Then it is very easy to see that $\lfloor \sqrt{n!}\rfloor = k-1$, which divides $n!$. So proving your claim requires proving Brocard. – Erick Wong Feb 8 '17 at 20:53
• @ErickWong wrote out the argument in my answer. – Will Jagy Feb 8 '17 at 21:13

Added: nice argument at Finding the Number of Positive integers such that $\lfloor{\sqrt{n}\rfloor} \mid n$ where all numbers $m$ such that $\lfloor \sqrt m \rfloor | m$ are of the form $m = k^2$ or $m = k^2 + k$ or $m = k^2 + 2 k;$ furthermore, this is if and only if.

It is easy to show $n!$ is not a square by Bertrand's. If we could prove the current conjecture, that would be a proof that, for $n \geq 8,$ $$n! \neq k^2 + k$$ and $$n! \neq k^2 + 2k.$$ That is, we would have proved that $$4 n! + 1 \neq 4k^2 + 4k + 1,$$ $$n! + 1 \neq k^2 + 2k + 1,$$ indeed $$4 n! + 1 \neq v^2,$$ $$n! + 1 \neq w^2$$ for $n \geq 8.$ A proof of the current conjecture would include a proof of Brocard. This conjecture is stronger than Brocard. Also stronger than dirt.

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Well, see what can be proven out of this. Interesting for $n=8,$ there is no new prime showing up in "sqrt," rather the exponent of 5 is too large.

2  sqrt  1 =   1   fact  2 =  2
3  sqrt  2 =  2  fact  6 =  2 3
4  sqrt  4 =  2^2  fact  24 =  2^3 3
5  sqrt  10 =  2 5  fact  120 =  2^3 3 5
6  sqrt  26 =  2 13  fact  720 =  2^4 3^2 5
7  sqrt  70 =  2 5 7  fact  5040 =  2^4 3^2 5 7
8  sqrt  200 =  2^3 5^2  fact  40320 =  2^7 3^2 5 7
9  sqrt  602 =  2 7 43  fact  362880 =  2^7 3^4 5 7
10  sqrt  1904 =  2^4 7 17  fact  3628800 =  2^8 3^4 5^2 7
11  sqrt  6317 =  6317  fact  39916800 =  2^8 3^4 5^2 7 11
12  sqrt  21886 =  2 31 353  fact  479001600 =  2^10 3^5 5^2 7 11
13  sqrt  78911 =  7 11273  fact  6227020800 =  2^10 3^5 5^2 7 11 13
14  sqrt  295259 =  295259  fact  87178291200 =  2^11 3^5 5^2 7^2 11 13
15  sqrt  1143535 =  5 228707  fact  1307674368000 =  2^11 3^6 5^3 7^2 11 13
16  sqrt  4574143 =  7 31 107 197  fact  20922789888000 =  2^15 3^6 5^3 7^2 11 13
17  sqrt  18859677 =  3 37 131 1297  fact  355687428096000 =  2^15 3^6 5^3 7^2 11 13 17
18  sqrt  80014834 =  2 79 506423  fact  6402373705728000 =  2^16 3^8 5^3 7^2 11 13 17
19  sqrt  348776576 =  2^7 139 19603  fact  121645100408832000 =  2^16 3^8 5^3 7^2 11 13 17 19
20  sqrt  1559776268 =  2^2 139 2805353  fact  2432902008176640000 =  2^18 3^8 5^4 7^2 11 13 17 19
21  sqrt  7147792818 =  2 3^2 19 20899979  fact  51090942171709440000 =  2^18 3^9 5^4 7^3 11 13 17 19
22  sqrt  33526120082 =  2 7 13 184209451  fact  1124000727777607680000 =  2^19 3^9 5^4 7^3 11^2 13 17 19

• The anomaly with $n=8$: a consequence of flooring? – Maazul Feb 8 '17 at 18:34
• The other case $n! = k^2+k$ was previously asked here math.stackexchange.com/questions/350637/…. It appears the OP's claim that this can be easily eliminated is not credible. – Erick Wong Feb 8 '17 at 23:57
• @ErickWong evidently I answered that one, with information from Guy's book... I remember, getting ready for being a TA, we were told to think of mathematics as another language. This is a case where it seems clearer to write things out in symbols, and where the OP keeps arguing in English, without necessarily having written anything down carefully. – Will Jagy Feb 9 '17 at 0:49
• @ErickWong Thank you for explaining why we cannot rule out the case $n!=k(k+1)$. I can see how this is stronger than Brocard. – Maazul Feb 9 '17 at 3:08