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)
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.
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.
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
I typed a similar equation, "n! = n^2 + 2n", on Wolfram Alpha, and it gave two numerical solutions but didn't give the more obvious solution (you can guess what that is).
![KSI - Dirty [Official Visualiser]](http://i.ytimg.com/vi/VoDh1h1dPUE/mqdefault.jpg)
The interesting thing about this problem is that it relates back to a problem that Prime Newtons did a few days back
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)
Very neat solution!
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.
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
What a lovely video today!
😁
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.
Nice solution in the video!
120+6=126
Yes! I have done it 🎉
I also got n=5.
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
I typed a similar equation, "n! = n^2 + 2n", on Wolfram Alpha, and it gave two numerical solutions but didn't give the more obvious solution (you can guess what that is).
5
Ez if u know sum of cubes is just that multiplication great video!
Spell out your words.
(n-3)! = 1+ 3/(n-2). To make 3/(n-2) = integer, n=5