A Factorial Diophantine Equation (n^2+19n-n!=0)

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[SyberMath]

Dear friends, here is two videos on factorial equations:

[guisav]

One way to limit the guessing game and to prove there are no more answers is to notice that, if k+20=k!, and k is not zero, then, dividing both sides by k, 1+ 20/k = (k-1)!. Since (k-1)! is an integer, so is 20/k, and then k divides 20. So we only have to check 1, 2, 4, 5, 10 and 20. Only 4 works.

[littlefermat]

That's why inequalities are important not just for algebra, but number theory as well!

[XJWill]

If we use the usual definition for the gamma function, then we can substitute
n! = gamma(n+1)
into the equation. The only integer solution to n^2 + 19*n - gamma(n+1) = 0 is n=5, but there are several real-valued solutions:
n ~ 0.051
n ~ -1.0546681
n ~ -1.97
n ~ -3.01
.
.
.

[raystinger]

So the first thing I noticed from the original equation is that it has a common factor of n, and assuming n>0, we can divide this common factor:
n² + 19n - n! = 0
=> n + 19 - (n-1)! = 0
=> n + 19 = (n-1)!
So, in the left hand side we have a simple linear function and on the right hand side we have a factorial. Given that factorials increase in a much faster rate than linear functions, therefore n must be a small value. So we can just check some values untill we get an answer:
n = 1, n+19 = 20, (n-1)! = 1
n = 2, n+19 = 21, (n-1)! = 1
n = 3, n+19 = 22, (n-1)! = 2
n = 4, n+19 = 23, (n-1)! = 6
n = 5, n+19 = 24, (n-1)! = 24 => This is an answer!
n = 6, n+19 = 26, (n-1)! = 120
n = 7, n+19 = 27, (n-1)! = 720
As we can see, n=5 is an answer. Also, for n>5, (n-1)!>n+19, so there aren't any more integer solutions for this equation.
Answer: n=5

[-basicmaths]

Above equation We can write as
n+19/(n-1)(n-2) =(n-3)!
Here n is odd because the product ( n-1)*(n -2) is even. Secondly numerator have less power than denametar n have one of the least values. n=5 satisfied with equation.

[knutthompson]

Factorial functions are often not that interesting. Factorials and polynomials, and even exponentials (with a constant base anyway), don't play well together since factorials grow so quickly, so there is pretty much never more than one answer, and it is usually pretty small. Since they are integers, you can guess an answer without too much trouble. Then quickly conclude it is the only answer based on growth rate. So it boils down to a guess and check. Which is fine for what it is worth, but like I say, somehow not that satisfying. I do always enjoy your channel, though. :-)

[soufianeaitabbou]

By Wilson's theorem we can easily figure it out, since n+19=( n_1)!
by the Wilson's theorem : (n-1)!=_1 mod n
so n+19 = _1 mod n : n+20 = 0 mod n
wich means : 20 = 0 mod n so n divides 20
thus n={1, 2, 4, 5, 10, 20} after checking cases : n=5

[italixgaming]

My solution (of course way faster):
n²+19n=n!
n=0 doesn't work and if n>0, n²+19n >= 20>3! then n>3
So we can divide everything by n(n-1) and still have integers:
(n+19)/(n-1)=(n-2)! <=> 1+20/(n-1)=(n-2)!
Since n>3, (n-2)! is an even number then 20/(n-1) is an odd number, which also divides 20.
But knowing that 20=2².5, the only odd divisors of 20 are 1 and 5.
1 can't be a solution since 1+1=2 is not a factorial but 5 works: 1+5=6=3!
Then n-2=3 then n=5.

[stevenwilson]

I did some playing around and you can factor the system into: n(n + 19 - (n-1)!) = 0, and since n = 0 is not a solution (which you can easily check), this becomes n +19 = (n-1)!, or more generally 19 = (n-1)! - n; thus for any system, b = (n-1)! - n. for 19, this is 4! - 5, so n = 5. But you can use this to solve for any n, given b by knowing the pattern of the factorials, 1, 1, 2, 6, 24, 120, … etc.

[math]

love this tutorial on factorial equations, thanks for sharing

[IvyANguyen]

Thanks for the video showing how to do it. I tried to do it via guess & check whilst seeing the thumbnail and arrived at n = 5 after trying a few numbers out for almost the same time this video lasts. Knew straight away 0 won't work because 0!=1.

[capjus]

You see in this video you start directly, thx for having listened to me :)

[albinbackstrom]

Another thing I thought of was that you can rearrange the equation k+20=k! into 20=k!-k. This can then be factored as 20=k((k-1)!-1). This means that both k and (k-1)!-1 is a factor of 20. It is obviously a really easy equation at this point but it is always funny to continue. Now considering (n-1)! is always even at n>2 (n=2 and anything lower gives us (2-1)!-1 or less which is 0 or less which obviously is not a factor of 20) we can deduce that (n-1)!-1 is odd which means it is either 1 or 5. 1 is quite obviously impossible considering 19! is an absolutely huge number. Therefore (n-1)!-1=5 and n=4. Unecessary steps but I thought it was interesting nontheless.

[mvrpatnaik]

The way the problem is solved is very impressive

[diedoktor]

I figured guess and check would be the best way to solve this, I found n=5 is a solution pretty quickly. The factorial part will grow faster than the quadratic part so I believe that's the only solution.

[benardolivier]

n² + 19n = n! -> n(n+19) = n! -> n+19=(n-1)! -> (n-1)+20=(n-1)! -> (n-1)! - (n-1) = 20 -> (n-1)*((n-2)! - 1) = 20 -> n-1 divides 20, then just check manually for n-1 = 1, 2, 4, 5, 10, 20 and you're done (works only for n-1=4, or n=5).

[pierretoulouse]

you can get (n-1)! - (n+18) =1, then according to Bezout theorem, only possible solutions are (n-1)! and (n+18) are prime to another, so you set array with values from 1 to 5 and eliminates couple not prime to each other and it rests n=5.

[rukukaru]

My brain is currently being fried, why did I watch this when I'm feeling a fever

[mbmillermo]

I think the quickest way to deal with this kind of problem is to understand that n! increases very fast with increasing n, so n has to be a small number. Try something and if it is too big or too small, try again. You'll get the answer almost immediately -- no need for algebra or any real thinking.