Let's Solve A Nice Factorial Equation

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PLAYLISTS 🎵 :

[JourneyThroughMath]

The interesting thing about this problem is that it relates back to a problem that Prime Newtons did a few days back

[dugong]

If you ask WA "n!+(n-2)!=n^3+1 ; n is an integer" it does give the solution but still doesn't show the part of the graph where the curves intersect (n=5)

[NowInAus]

What a lovely video today! <—- not a factorial symbol 😂

[toveirenestrand]

n! + (n - 2)! = [n(n - 1) +1](n - 2)! = (n^2 - n + 1)(n - 2)! =n^3 + 1. n taking on the values 2, 3 and 4 does not fulfill the equation, but n = 5 does: (25 - 5 + 1)3! = 21*6 = 126,
which equals 125 + 1 = 126. So n = 5.

[FisicTrapella]

As you, I'll got (n-2)! (n^2 - n + 1) = n^3 + 1
In this point, notice that n must be >= 2 and 3<=n<=5 to have a cube as in the RHS.
Just check 3, 4 and 5 and you'll see that only n=5 works. 🤓

[goldfing]

n = 5 fulfills the equation, since
5! + 3! = 120 + 6 = 126 and
5^3 + 1 = 125 + 1 = 126, too.
I double any greater solution cos the faculty grows more quickly than cubing.

[lawrencejelsma]

A great video. How did Wolfram Alpha get as lost as my brain? Obviously n=1.374354 ... doesn't have a factoral like the factorials of negative useful solutions. Like your solution I was only removing until (n-2)! on the left hand side and didn't realize the n^2 - n + 1 terms that only cancel if not imaginary ns. That was clever using the k substitution to move further along and subtracting k I wouldn't have thought of. When you got k=3 as a solution and we checked n <= 2 fails. k=n-2=3 only remaining result was surprising to have n=5 solution that Wolfram Alpha was getting graphically by not figuring out (n-2)! part properly, I think!? 😂

[danielc.martin]

Ez if u know sum of cubes is just that multiplication great video!

[feiyuqiu]

(n-3)! = 1+ 3/(n-2). To make 3/(n-2) = integer, n=5