$n$ is an integer greater than $7$. How does one go about proving that $\lfloor \sqrt{n!}\rfloor\nmid n!$.
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3Relevant: http://math.stackexchange.com/q/6369/73324 – vadim123 Feb 08 '17 at 18:12
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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 08 '17 at 18:15
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1Also: http://math.stackexchange.com/questions/12544/ – i9Fn Feb 08 '17 at 18:26
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This would imply Brocard's conjecture, so certainly no proof is yet known. – Erick Wong Feb 08 '17 at 19:03
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The third case where $n! = k(k+1)$ can probably be attacked using abc conjecture, and I would guess it is also open otherwise. – Erick Wong Feb 08 '17 at 19:09
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This is actually weaker than Brocard's Problem. An additional condition $\lfloor \sqrt{n!}\rfloor -1 \nmid n!$ for $n>7$ would be required for Brocard. – Maazul Feb 08 '17 at 19:09
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If $n!=k(k+1)$ then one of the factors is even and the other is odd. But for $n>7$, they cannot differ by one because the odd factor will be the product of consecutive odd numbers and the even factor will be the product of consecutive even numbers. So this case is easily out of question. – Maazul Feb 08 '17 at 19:11
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@Maazul In that case your question is trivially equivalent to Brocard. But I'd like to hear the argument for $k(k+1)$, it didn't leap out as obvious to me. – Erick Wong Feb 08 '17 at 19:16
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If you imply that proving this would solve Brocard, then no. For Brocard, it would require in addition to prove that $\lfloor \sqrt{n!} \rfloor - 1 \nmid n!$ for $n>7$. – Maazul Feb 08 '17 at 19:20
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@Maazul Then I don't follow your earlier comment. The condition is equivalent to $n!$ belonging to one of the forms $k^2, k(k+1), k^2-1$. The first is easily eliminated, the second you claimed to be easy, and the third is Brocard. – Erick Wong Feb 08 '17 at 19:29
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@ErickWong Proof of Brocard would certainly prove the statement in question. However, the converse is not true. You see, even if proven, you have to rule out the additional condition by virtue of Brocard requiring a gap of 2 in the factors. – Maazul Feb 08 '17 at 19:37
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@Maazul Brocard's problem concerns the existence of solutions to $n! + 1 = k^2$. What additional gap requirement are you referring to?? – Erick Wong Feb 08 '17 at 19:45
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$n!=k^2-1=(k+1)(k-1)$. If the factors of $n!$ differ by 2, then the equation holds. Conditions $\lfloor \sqrt{n!} \rfloor -1\nmid n!$ and $\lfloor \sqrt{n!} \rfloor \nmid n!$ for $n>7$ are required to be proven otherwise. – Maazul Feb 08 '17 at 19:50
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1@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 08 '17 at 20:53
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1@ErickWong wrote out the argument in my answer. – Will Jagy Feb 08 '17 at 21:13
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Possible duplicate of Finding the Number of Positive integers such that $\lfloor{\sqrt{n}\rfloor} \mid n$ – Ross Millikan Feb 08 '17 at 21:17
1 Answers
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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
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The other case $n! = k^2+k$ was previously asked here https://math.stackexchange.com/questions/350637/find-the-positive-integer-solutions-of-m-nn1. It appears the OP's claim that this can be easily eliminated is not credible. – Erick Wong Feb 08 '17 at 23:57
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1@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 09 '17 at 00:49
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@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 09 '17 at 03:08