a nice factorial equation.

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[xoriun]

There is a much simpler approach: subtract 1 on both sides, so you have a difference of squares on the right side, cancel the (n-1) term and you are left with the following:
(n-2)! = n+1 which is really easy to check since the right side only increases linearly, so once the left side is larger, your done. (n=1 has to be checked with the original equation though)

[thatdude_]

An easier way: check the case n=1 and n=2 first to see they don't give solutions, then pull the +1 to the other side so the equation reads (n-1)!=n²-1. Since this factors nicely we have (n-1)!=(n-1)(n+1). Since we assume n>2, we can cancel and get (n-2)!=n+1. Trying some values, we find that only n=5 seems to give a solution, and for n>5 (n-2)! grows much larger than n+1. So all we're left with is to prove that (n-2)!>n+1 for all n>5, which can be done quickly by induction. Hence n=5 is the only solution.

[Kapomafioso]

How about doing this instead:
(n-1)! + 1 = n^2
(n-1)! = n^2 - 1 = (n-1)(n+1)

Here, n >= 1 obviously. If n = 1, you get
0! = 1 = 0*2 = 0, so n = 1 isn't a solution

If n > 1, you can divide out n-1 and get
(n-2)! = n+1

The only solution is n = 5, to prove that there is no solution for n > 5, you notice that at best, while the right hand side increases by 1 by increasing n, the left-hand side gets multiplied by 3.

[zanti]

At times Prof. Penn likes to use nuclear weapons to kill bugs. This video, for example. My idea is to note that 5! + 1 = 11², and since 11 - 5 = 6 > 1, that establishes 5 is already an upper limit for the possible solutions.

[Halfjera]

another quick obs here - n can only be odd, because n! is almost always even, so n!+1 mod 2 is always 1. Thus, any perfect square on the right must also be odd, and that only happens when n is odd. You can easily see that 7 would never work, 3 doesn't work either, so you're left with 5

[mcwulf]

(n-1)! = n^2 - 1 = (n+1)(n-1)
Cancel (n-1)
(n-2)! = n+1

There comes a value of n when LHS> RHS,
(n+2)! >= (n-2)(n-3) > (n+1)
Leads to
n^2 - 6n - 5 > 0
(n-5)(n-1)>0
n > 5

So we only need to check n <= 5.

[laurentpaget]

(n-1)!+1=n^2 <=> (n-1)!=n^2-1=(n+1).(n-1). Dividing both sides by n-1 (n>1), the equation becomes (n-2)!=n+1. F(n)=n+1 is a linear function, whereas (n-2)! is an exponential function. There is only one solution, as both functions are monotonically increasing. Checking for n=0, 1, 2, … shows that n=5 is the only solution.

[michaelslack]

I think it's easier if you remove a factor of n-1 and then compare factorial to linear growth... Unless I'm missing something

[Oscar]

I like the (n-2)!=n+1 method more, but I would also like to add that n must be prime:
if p<n is a prime, then p|(n-1)! and therefore is coprime with (n-1)+1=n^2, so n can't be a composite number otherwise it would be coprime with one of its prime divisors.

[drsonaligupta]

Hands were itching to take the 1 to the other Side and Kill the (n-1).
(Assuming n>1 as n=1 doesn't work)

[arandomcube]

I did it quite a different way.
I fistly noticed 5 was the only solution under 6, then showed that (n-1)!+1 was strictly bigger than n^2 for n>5 by a reasonably simple induction. This intuition is simple : find a rank where the left side of the equation is superior to the left side, and then show it will always be the case from that rank, as the factorial is increasing way faster than any polynomial (of degree 2 in particular). Furthermore, factorials and polynomials are objects that are usually not too hard to deal with in inductions.

The base case is obvious ((6-1)!+1 = 121 > 36). Let's consider it's true for n>5 and then show it is too for n+1.
The induction hypothesis is :
(n-1)!+1 > n^2. Multiply by n and then substract n-1 on both sides, we get :
n! +1 > n^3 -n +1. We just have to show that this is bigger than (n+1)^2. This is how I did it. As it is a quite simple inequality, there is probably zillions of way to show it, and probably many that are more simple than mine. Anyway :
n^3 = n * n^2, but since n>5, we have :
n^3 > n^2 + n^2+ n^2 + n^2 and then n^3 -n +1 > n^2 +(n^2-n+1) + n^2 + n^2.
But if n>5, we obviously have : n^2-n+1 > n^2-2n+1 = (n-1)^2 >= 0, n^2 > 2n, n^2 > 1.
This finally gives us : n^3 -n+1 >= n^2 +0 +2n +1 = (n+1)^2, which finishes up the proof.
5 is definitly the only solution.
And that's a good place to end this comment :)

[skylardeslypere]

I thought I was smart by finding the solution on my own by dividing out a (n-1) term, but when I came to the comments I saw that pretty much everyone had this idea too

[davidcroft]

Ah! I like so much this type of problem

[mathphysicsnerd]

This is definitely one of those "killing a flea with a sledgehammer" solutions

[malabikasaha]

Speaker's approach takes in guesses as well. Just note that the problem reduces to (n-2)! =(2/n)(sum of first natural numbers). So the starting guess is 1. And it converges at 5 alone.

[jaimeduncan]

The inequality: for people that don't know calculus one can factor the cubic and get (m-3)*(m+1)^2. and use high school math to solve it. The conclusion is the same, for. m>3 it's always positive, but since we are in the natural number it means m>=4 same conclusion.

[wesleydeng]

Michael has so many weapons in his arsenal that he always has a tough time to choose which one to use.🤣

[TheAdwatson]

I used Open Office Calc and found n=5 in less than the time it took to input the equation. Since factorials only apply to integers, sometimes a "brute force" computer approach is far quicker than an analytical one, if you only want to find the answer to the question, rather than why.

[sugarfrosted]

You can quickly see the solutions must be 1 or prime using modular arithmetic.

[themanwhocarriesthesun]

Since n is a natural number and you can check that for n= 1 to 3, they do not satisfy the equation (n-1)! + 1 = n^2(*). That means n has to be greater than 3. Since n>3, (n-1)! has to be even. Thus, (n-1)! + 1 and n^2 are odd. Thus, n is odd. Since n is odd and n>3, we must have n>=5.

Now let's write equation (*) as
(n-1)! = n^2 - 1 = (n-1)(n+1)
(n-2)! = n+1 = n-2 + 3
(n-2)[(n-3)! - 1] = 3.

From the condition n>=5, we have
n-3 >= 2
(n-3)! >= 2! = 2
(n-3)! - 1 >= 1
(n-2)[(n-3)! - 1] >= n-2 >= 3.

It means that the only solution that satisfies equation (*) is n=5.