Solving an Intriguing Diophantine Equation with Factorials | Number Theory Challenge

Solving an Intriguing Diophantine Equation with Factorials | Number Theory Challenge
Discover the fascinating world of number theory as we tackle an intriguing Diophantine equation with factorials in this math challenge! In this video, we'll dive deep into the concepts of Diophantine equations and factorials, exploring their unique properties and applications in mathematics. Join us on this exciting journey as we break down the problem step by step, providing you with the tools and insights needed to solve this captivating number theory challenge. Whether you're a math enthusiast or just looking to sharpen your problem-solving skills, this video is sure to pique your interest and expand your knowledge. Don't miss out on this thrilling mathematical adventure!

#diophantineequations #factorials #numbertheory #mathchallenge #algebra #mathematics #cubicequations #factorialnotation #factorial #math #integer

Topics Covered:
1. Diophantine Equation
2. Factorials
3. Solving Cubic equation with algebraic identities and manipulations
4. Number Theory Tips
5. Solving cubic equation
6. Finding solutions

7 Key moments of this video:
0:00 Introduction
0:21 What is factorial?
1:38 Solving Diophantine equation
2:22 Difference of squares
2:51 Substitution
5:07 Solving for y and n
7:16 Verify Solution

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[devondevon]

Answer = 5
n^3 = n + n!
n^3 - n = n! substract n from both sides
n(n^2 - 1) = n (n-1)! factor out n on the left and right sides
n^2 -1 = (n-1)! divide both sides by n
(n+1)(n-1) = (n-1)! factor the left side
= (n-1)(n-2)!
n+1 = (n-2)!
n is greater than 3 since 1! is 1 since +1 is on the left side, n is also odd since n+1 makes
it even and factorial are even except 0!, but can not = 0 given (n-2) hence =5
5 +1 = (5-2)!
6 = 3!

[ANGELOMAGDAONG-cq]

there's no need to do y substitute, transpose n, we will have n^3 - n = n! . n(n^2-1) =n!, n(n+1)(n-1) = n!, divide both sides by n(n-1), n+1 = (n-2)! from here we can do some trial and error, so n=5. as n approaches infinity from 5, the right side is increasing much faster, so the only solution is n=5

[walterwen]

Number Theory Challenge: n^3 = n + n!; n = ?
n! = n^3 – n; n is a positive integer > 1
n! = n(n – 1)(n – 2)! = n^3 – n = n(n – 1)(n + 1), (n – 2)! = n + 1; 6 > n > 4
x = 5; (n – 2)! = 3! = (3)(2)(1) = 6 = 5 + 1 = x + 1; Confirmed
Final answer:
n = 5

[brianxx]

Why do a page or two of algebra and then brute force a solution? You could have more easily tried (brute forced) n = 0 up to n = 5. No algebra needed, and no arithmetic more advanced than 5^3 and 5!

0 = 0+1 no.
1 = 1+1 no.
8 = 2+2 no.
27 = 3+6 no.
64 = 4+24 no.
125 = 5+120 yes. Done.

I assume everyone here can multiply by 5 with less effort than implementing even the simplest substitution. What was y for?? What did it gain you?



To those looking for proof of no n over 5:
Moving from n=5 to n=6, the left side expands by 1.2^3 while the right side expands by nearly a factor of 6. From there, the left side grows more and more slowly while the right side grows more and more quickly. No higher n is possible.

[italixgaming]

Jeeez you completely overcomplicated this poor little thing. Here is how to destroy this.
We rewrite the equation as this: n^3-n=n!
But n^3-n=n(n²-1)=(n-1)n(n+1)
Since 0 and 1 are not solutions of the equation (the LGS is equal to 0 but a factorial is always strictly positive), we can put n(n-1) in factor of both sides and divide by it (and keep the equivalence). We obtain: n+1=(n-2)!
Now we can quickly see that 2, 3 and 4 are not solutions of the equation, simply because 3, 4 and 5 aren't factorials of any number (the lowest factorial after 2!=2 is equal to 3!=6)
Therefore n>4 and (n-2)! can be written as the product of (n-2) and a set of integers where 2 is present (since 2<n-2). Which means that, for n>4, (n-2)! is greater or equal to 2(n-2).
Now we see that 2(n-2)-(n+1)=n-5>0 for n>5, which implies that 2(n-2)>n+1 for n>5, which means that (n-2)!>n+1 for n>5.
So we have only candidate left, which is n=5.
Let's verify: n+1=6 and (n-2)!=3!=6.
Conclusion: our original equation has only one solution, which is 5.

[davidbrisbane]

You also need to show that there are no more possible solution other than n = 5.

[danielderoudilhes]

Sorry, but is not a demonstration. It's just successif trying.