n^3 = (n+2)!!

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@chaosredefined3834 5 ай бұрын

n^3 = (n + 2)!! Therefore, n^3 = (n + 2) x an integer. Note that n^3 + 8 = (n + 2)(n^2 - 2n + 4). So, n^3 + 8 = (n + 2) x an integer. From this, we can see that 8 = (n + 2) x an integer. Therefore, n + 2 is either 1, 2, 4 or 8. So, n is -1, 0, 2 or 6. (-1)^3 is -1, and (-1 + 2)!! = 1, these aren't equal, so n=-1 is not a solution. (0)^3 is 0, and (0 + 2)!! = 2, these aren't equal, so n=0 is not a solution. 2^3 = 8, and (2 + 2)!! = 4!! = 8. These are equal, so n=2 is a solution. 6^3 = 216, and (6 + 2)!! = 384, these aren't equal, so n=8 isn't a solution. Thus the only solution is n=2.

@prateekbhurkay9376 5 ай бұрын

Your solution was way better

@chaosredefined3834 5 ай бұрын

@@prateekbhurkay9376 Thanks. This was roughly the plan I was expecting the video to use, and was kinda surprised he didn't use it.

@voicutudor7331 5 ай бұрын

great solution, he just threw some examples around and said he solved it 😂

@donovanjimmyvasquezmolinar9648 5 ай бұрын

Sorry for nitpicking, but how do you know that 8 = (n+2) × an integer from n^3 + 8 = (n+2) × an integer?

@voicutudor7331 5 ай бұрын

@저녘놀 5 ай бұрын

I thought that n!! = (n!)! at first, but it was not.

@display9138 5 ай бұрын

Same lol

@dembotrawally9431 5 ай бұрын

You're just great.. keep it up.. I like your methodology

@Ron_DeForest 5 ай бұрын

I’ve seen enough of your videos to find it difficult to believe you can’t answer any math question posed to you.

@PrimeNewtons 5 ай бұрын

I am laughing really hard 🤣. KZbin has a way of making us look smarter than we really are. I appreciate your kind words.

@atharvsharma5111 5 ай бұрын

he aint forcing you to watch his vedios

@filipeoliveira7001 5 ай бұрын

Bro hasn’t seen the IMO😭🙏

@dontmind159 5 ай бұрын

@@filipeoliveira7001 what's that about?

@santoshpanjiara1860 5 ай бұрын

He didnt insult him. He instead praised him. Read the comment once again. Learn english brother.@@atharvsharma5111

@RubyPiec 5 ай бұрын

can you solve all 7 millennium problems in the next video? thanks

@Jamesdaman69 5 ай бұрын

7 million dollars right there 😂

@rivenoak 5 ай бұрын

he could show us the problems for sure and tell us why those are unsolved at the moment.

@YakobWakjera 5 ай бұрын

@@rivenoaklove it

@martintuma9974 5 ай бұрын

@@Jamesdaman69 6 million, one of those problems has been solved already.

@PrimeNewtons 5 ай бұрын

Let's goooooo 🤣🤣🤣🤣🤣

@nathannibbs 5 ай бұрын

You've been my secret weapon to learn math through my educational career so far. Never stop these vids

@אסףרביבו-ר7ת 5 ай бұрын

Your a great techar well done man

@hiteshliya8639 3 ай бұрын

Very knowledgeable channel!!! Thank you for the knowledge

@BinaHejazi 5 ай бұрын

well explained and well delivered, thoroughly enjoyed it mate!!

@TuhafOyuncular 5 ай бұрын

n^3=(n+2)!! n^3=(n+2)*(n)*(n-2)!! n^2=(n+2)*(n-2)!! 4/(n+2)+n-2=(n-2)! 4/(n+2) is an integer: n+2=4,2,1,-1,-2,-4 n=2,0,-1,-3,-4,-6 n+2>0 and (n+2)!!>0: n>-2 and n>0 n>0 n=2,0,-1,-3,-4,-6 and n>0: n=2 Therefore answer is only 2 for integers.

@mab9316 5 ай бұрын

Best and most complete solution. Thanks.

@Moss-q9o 5 ай бұрын

What happened from the 3rd to the 4th line?

@TuhafOyuncular 5 ай бұрын

@@Moss-q9o In 3rd line, we use some basic rules: Number in double integral's value must be positive integer and double integral's value must be positive number. In 4th line, we use 2nd line's n values and check them for 3rd line's conditions for solutions.

@Moss-q9o 5 ай бұрын

@@TuhafOyuncular i meant how did we get to 4/(n+2)+n-2=(n-2)!

@ferderka 5 ай бұрын

@@Moss-q9o he divided x^2 by (x+2) and got 4/(x+2) +x-2. Also it looks like he forgot to type second exclamation mark

@ethanbartiromo2888 5 ай бұрын

You can prove this even more rigorously by showing that it will never overtake again after 7 for odd numbers because (n+2)*n > n^2 and (n - 2) * 3 >= n for all odd n >= 3, but it only applies in the double factorial when n - 2 > 3, when n - 2 > 3, so then multiply each side, (n + 2)!! >= (n + 2)*n*(n-2)*3 > n^2 * n = n^3 for all odd n >= 7, similarly, this applies to all even n >= 6, because (n + 2)*n > n^2 and (n-2)*2 > n for all even n >= 4, but 2 is not listed twice is double factorial, so it must be even where n > 4, so it will never increase past the double factorial again. However, if we look at negative values of n then one side will always be negative and the other will always be positive, or not defined, so it would be cool to look at |n^3| = |n + 2|!! Which actually has more solutions, notably n = -1 and the solutions would effectively be the same as the negation of the solutions to n^3 = (n - 2)!! For positive n, which we can find that at the worst case once (n-2)(n-4)(n-6)(n-8) > n^3 it will always increase faster, which I’ll leave it to the reader to find any other solutions if they do exist

@dmytryk7887 5 ай бұрын

Another way. Divide both sides by n giving n^2=(n+2)(n-2)!! Then write the left side as n^2-4+4. Divide both sides by n+2 and the left side becomes n-2 +4/(n+2). Since this must be an integer the only possible value for n is n=2. Check and it works.

@shade4220 4 ай бұрын

thats a really cool method. kudos!

@hazalouldi7130 4 ай бұрын

very nice that's a proove.

@lukashilebrand 5 ай бұрын

it's so fun to watch your videos. Thanks!!

@neytirisbest 5 ай бұрын

Hey, thank you for your brilliant clips! Though (or because) being a physicist, math was never my favorite subject. But your clips are -in the meantime- kind of ritual and intellectual breakfast TV for me. Though I LOL sometimes think: who the hell exactly needs that… like double-factoriel….😂 Many greets from Germany 👍

@doctorb9264 5 ай бұрын

Cool problem and good analysis/solution.

@meaningwhat8371 Ай бұрын

Everytime that I watch his video, I am amazed at his chalkboard writing occupying the space appropriately.

@AyoSword 5 ай бұрын

It simply just comes down to the fundamentals and cool properties of 2, which is probably what inspired the problem. n ^ 3 = is just n x n x n, meaning there's no other factor other than n of course. This already tells you, that n + 2 = n x b, b = some integer. However, since n^3 has no factors other than n, b = n ^ c, c = some other integer n + 2 = n x n ^ c, n =/= 1 since n + 2 and n are both factors of n ^ 3. But then, (n + 2)!! = (n + 2) x n x (n - 2)!! = n ^ c x n x (n - 2)!! = n ^ (c + 1) x (n - 2)!! = n ^ 3 (c + 1) must be less than or equal 2 to make up for the (n - 2)!!. And then again, n + 2 = n x b and n = n x 1, b < 1 since n < n + 2, so c =/= 0. b also must be positive as negative numbers have -1 as a factor, so: 0

@labmonkeyvimes42 3 ай бұрын

I think it can be observed from the onset that n must be even since if n is odd n^3 will be a unique combination of numbers that cannot be replicated on the other side of (n+2)!! and you will have prime numbers in the breakdown that cant overlap.

@nanamacapagal8342 5 ай бұрын

My take: Split into the odd case and the even case. EVEN CASE: -2!! doesn't exist, so n > -4. Lower bound done, now for the upper bound. 8!! = 384 6^3 = 216 For all even n greater than 6, (n+2)!! > n^3. PF: Assume n is even, n > 6 and n^3 > (n-2)!! (n+2)!! = (n+2)n!! > (n+2)(n-2)^3 = n^4 - 4n^3 + 16n - 16 > 6n^3 - 4n^3 + 96 - 16 = 2n^3 + 80 > n^3 We are bounded with -4 < n < 6 Now check all possible n since the sample space is small: n = -2: n^3 = -8, (n+2)!! = 1 n = 0: n^3 = 0, (n+2)!! = 2 n = 2: n^3 = 8, (n+2)!! = 8 n = 4: n^3 = 64, (n+2)!! = 48 We have the solution n = 2. ODD CASE: Similar argument. -5!! is not an integer, so n > -7. 9!! = 945 7^3 = 343 For all odd n greater than 7, (n+2)!! > n^3 PF: Assume n is even, n > 7 and n^3 > (n-2)!! (n+2)!! = (n+2)n!! > (n+2)(n-2)^3 = n^4 - 4n^3 + 16n - 16 > 7n^3 - 4n^3 + 112 - 16 = 3n^3 + 96 > n^3 Therefore for odd n, -7 < n < 7. We can check all cases n = -5: n^3 = -125, (n+2)!! = -1 n = -3: n^3 = -27, (n+2)!! = 1 n = -1: n^3 = -1, (n+2)!! = 1 n = 1: n^3 = 1, (n+2)!! = 3 n = 3: n^3 = 27, (n+2)!! = 15 n = 5: n^3 = 125, (n+2)!! = 105 No solutions when n is odd. Therefore the only solution in the integers is n = 2

@dronevluchten 5 ай бұрын

The best proof that n=2 is the only solution is that for n>2 other prime factors pop up in (n+2)!! which are not in n to the power 3.

@wira2562 5 ай бұрын

New knowledge, sir!🎉🎉🎉

@danbaker300 Ай бұрын

There's a fairly straightforward proof that there are no solutions at n >= 4. In that case, the equation can be rewritten as: (with 0!! = 1) n^3 = (n+2) * n * (n-2) * (n-4)!! Rearranging terms gives: (n-4)!! = n^2 / (n^2-4) = 1 + 4/(n^2-4) But for n >= 4, n^2-4 > 4 and therefore the right side is a non-integer value between 1 and 2. But the double factorial can only produce integers. So no solution can exist in that range. From there, it's easy to test n=1, 2, and 3 and verify that n=2 is the only solution.

@Christian_Martel 3 ай бұрын

11:22 “It’s not coming back”. No Coach, we’re not coming back!

@thaerthaer1120 5 ай бұрын

Are there any real applications for such equations in physics or other science

@shacharh5470 5 ай бұрын

if we assume n is even: n = 2m (2m)^3 = (2m+2)!! 8m^3 = (2m+2)!! 8m^3 = (2m+2) * (2m)!! = (2m + 2) * 2m * (2m - 2) * (2m - 4)!! 8m^3 = 8 * (m + 1) * m * (m - 1) * (2m - 4)!! m^3 = (m + 1) * m * (m - 1) * (2m - 4)!! m^2 = (m + 1) * (m - 1) * (2m - 4)!! m^2 = (m^2 - 1) * (2m - 4)!! so m^2 - 1 divides m^2. so m^2 = k * m^2 - k m^2 * (k - 1) = k notice k != 1 so we can divide by k-1: m^2 = k/(k-1) = 1 + 1/(k-1) 1/(k-1) is an integer so k-1 = 1 so k =2 so m^2 = 2 so m = sqrt(2) contradiction. What does it contradict? we assume we the expression (2m - 2) * (2m - 4)!! makes sense, but that's only true for m that's big enough, otherwise 2m-2 or 2m-4 is negative. That gives you an upper limit to n.

@SG49478 5 ай бұрын

I solved it with less trial and error. I wouldn't say it is much shorter though. If n^3=(n+2)!! and we have a product on both sides of the equation, both sides must contain the same prime factors. Therefor n+2 can not contain a prime factor, which n does not contain. Let P be the common prime factor product between n+2 and n. Then we can write n=k*P and n+2=l*P with n; l; k; and P all positive integers. If we subtract the first equation from the second, we get n+2-n=l*P-k*P. That means 2=P*(l-k). For positive integers there are two cases: case 1: P=1 and l-k=2. That means n+2 can have no other factor than 1, means n+2=1, n=-1 which is a contradiction. Therefor Case 2: P=2. That means n+2 contains only the prime factor 2 and therefor must be a power of 2. That also means n must be even. If n>2 the right side also contains the factor n-2. This one again can only contain primefactors that n contains as well. With the same reasoning as in regards to n+2 we can conclude that n-2 must be a power of 2 as well. Therefor n+2=2^a and n-2=2^b. a and b positive integers Subtracting both equations leads to 4=2^a-2^b. There is only one pair of powers of 2 with the difference of 4 which is a=3 and b=2. That would lead to n=6. We can test and quickly see that n=6 is not a solution. Therefor there is no solution for n>2. Since we established that n must be even the only choice left is n=2 which indeed is a solution and therefor the only solution.

@OlympicCode 5 ай бұрын

n + 2 divides RHS so it divides LHS too. Polynomial division for n^3 / n + 2 gives n + 2 divides 8. Which from there follows that n = 2 is the only solution.

@nozack5612 5 ай бұрын

I initially tried to solve with RHS = [ (n+2)! ] ! and concluded there are no integer solutions. The RHS gets very large much quicker than the LHS and the issue gets worse as n increases. However, looking at a plot of the gamma function, it appears there may be solutions somewhere among the reals (intersection of the appropriate gamma function and x^3).

@pojuantsalo3475 5 ай бұрын

If n>4, we have (n+2)!! = (n+2)(n)(n-2)(n-4)...p, where p is 1 or 2 depending on the parity of n. Multiplying the first three terms out gives (n+2)!! = f(n) = (n³-4n)*(n-4)!! Since 1!! = 1, we have f(5) = (5³-4*5)*1 = 105 < 5³ Since 2!! = 2, we have f(6) = (6³-4*6)*2 = 384 > 6³ It is pretty clear why f(n) > n³, when n > 5 since f(n) ≥ (n³-4n)*2 and it is easy to show the derivative of 2x³ is much greater than the derivative of 8x, when x > 5: g(x) = 2x³ ⟹ g'(x) = 6x² ⟹ g'(5) = 150 h(x) = 8x ⟹ h'(x) = 8 We can clearly see (n³-4n) > n³/2, when n > 5 which means f(n) ≥ n³, if n > 5 So, we only need to try out n = 0, 1, 2, 3, 4 and 5 (already done above) for solution. Doing that we find out n = 2 is the only solution.

@Psykolord1989 5 ай бұрын

Before viewing: Alright, first things to notice. Number 1, N must be a positive integer, aka a *Natural Number*. Why? Because 2 is a natural number, and we are able to take the factorial of (N+2), so N+2 must be a natural number as well. (Correction: You can check the cases of n = -1 and n=-2 if you like, but since the right hand side must remain positive and any negative cubed is a negative, you will find that neither n=-1 nor n=-2 are solutions) Number two: We are looking at double factorial. While you may be inclined to believe that double factorial means taking the factorial of a factorial, and thus getting some ridiculously large number, that 's not accurate. Rather, when taking double factorial, you treat it like a standard factorial, except that you only use the terms with the same parity (e.g. odd vs even) as the initial number. For example, 7! is (7)(6)(5)(4)(3)(2)(1) = 5040. However, 7!! = (7)(5)(3)(1) = 105. Similarly, 6! is 720, but 6!! = (6)(4)(2) = 48. Thus, double factorials are far smaller than regular factorials. So, we need to find an (n+2) such that (n+2)!! (that is, (n+2)(n)(n-2)(...) = n^3, I want to go at this through trial and error...mostly because that's something you can readily do with factorials, because you know you're dealing with natural numbers. Right off the bat, we know n=1 doesn't work (1^3 = 1, (1+2)!! = 3*1 = 3) However...n=2 gives us 2^3 = 8, and on the right hand side we have (2+2)!! = 4!! = 4*2 = 8 Thus, n=2 is a solution. Now, are there other solutions? Well, recall what I said above: (n+2)!! = (n+2)(n)(n-2)... If we have 3 terms there (which first happens at n=3, and happens again at n=4), our right hand side will come out to be n^3-4n. There does not exist a natural number for which n^3 = n^3 -4n, because it would mean that 0 = -4n, and 0 is not a "natural" number. The higher N goes, the more terms the right hand side has. At n=5 or n=6, we will have a n^4 term, and so on. As the number of factors on the right hand side increases, the *size* of those factors also increases. Thus, after n=2, the right hand side begins to outpace the left-hand side, and with each subsequent number that gap grows further and further. Thus, there will not be a chance for us to have a second solution. Our only solution, therefore, is *n=2*

@SakshamSiwa 5 ай бұрын

What did you do after highschool?

@herbertsusmann986 5 ай бұрын

I went through under grad math and physics and never saw a double factorial before. But I am 64 years old so maybe it is something they came up with since I was in University? Who uses double or triple factorials anyway?

@PrimeNewtons 5 ай бұрын

Differential equations

@ericongwae6742 4 ай бұрын

Although the solution will not be a natural number but is it possible to find the other non- integral value of n that lies between 5 and 6 because 5^3 is 125, (5+2)!! is 105 and 6^3 is 216, (6+2)!! is 384 hence the two functions must have crossed somewhere between 5 and 6.

@ACertainMan 5 ай бұрын

Eye balling the answer is 2 2³ = ( 2 + 2 ) !! = 4 * 2 = 8

@rivenoak 5 ай бұрын

in math often other solutions exist. but not today and you have to prove it

@okohsamuel314 5 ай бұрын

But 'eye balling' can be done by just about anyone...So it's not any special bro 😅

@ACertainMan 5 ай бұрын

@@okohsamuel314 Eye balling is a great skill to have.

@thaerthaer1120 5 ай бұрын

Can solve by Gama function ?

@PrimeNewtons 5 ай бұрын

Th8lis is double factorial. So I doubt it.

@OlavoLeopoldinodaSilvaFilho 4 ай бұрын

n³=(n²+2n)(n-2)!! (if n>2). Make a=(n-2)!!. Thus n²-an-2=0. Thus, the solutions for n must be such that their product must be -2, and sum must be a. But a must be natural. There is no such number. Then, we must have put n³=(n²+2n) (stopped short) -> n²-n-2=0->n=2 or n=-1. If n is natural, then n=2. Other way: (n-2)!!=1 or 0. Then n=3 (do not work) or n=2 (ok!).

@tcmxiyw 2 ай бұрын

You don’t need to multiply to check for equality-just look at the prime factors.

@johnconrardy8486 5 ай бұрын

just love your ividios

@scavengerethic 5 ай бұрын

hmm. if n^3 = (n+2)!!, then (n+2)!! must have no prime factors that n doesn't have, and the prime factorisation of the product of consecutive even numbers from n+2 down to 2 must contain only multiples of 3 of all the primes in it. If you divide by 2 until you reduce this product to the product of the first (n+2)/2 natural numbers, there's no way there are only cubes of all the 3s, 5s, 7s ... in that set. So the only way it works is if there is only one unique prime factor in n, and it has to be 2. So n+2 is a power of 2, so n = 2 and (n+2) = 4. There are probably gaps in my reasoning, I'm not a mathematician.

@objectifmaths 4 ай бұрын

if n>2, there exists a prime that do not divide both n and n+2 : only solution n=2. ok ?

@sayak6673 5 ай бұрын

Nice analogy

@gnanadesikansenthilnathan6750 5 ай бұрын

Got it. Easy problem

@mahoremujini 2 ай бұрын

On peut simplifier les tests en utilisant les divisibilités. Par exemple 4 n’est pas une solution car le membre de droite est un multiple de 6 et 4^3 ne l’est pas

@klausao 2 ай бұрын

n^3 = (n+2)!! = (n+2) * n * (n-2)*.... We can divide by n so we have n^2 = (n+2)*(n-2)!! But now if left side is a perfect square, the only solution is that n = 2 and (0)!! = 1.

@PrimeNewtons 2 ай бұрын

3³ is not 5!!

@klausao 2 ай бұрын

@@PrimeNewtons True, only n = 2 is the perfect square after divide both sides with n. Maybe I missed something :)

@surendrakverma555 5 ай бұрын

Thanks Sir

@ForestHills101 4 ай бұрын

For two term double factorial: n^3=(n+2)n, n(n+1)(n-2)=0 n=2 For three or more term double factorial: n^3=(n+2)n(n-2)m, m=>1 n^2=4m/(m-1) m=1: no solution m>1: 4m even, (m-1) odd, no solution n=2 only solution

@師太滅絕 5 ай бұрын

Unfortunately, solving by brute force is not a neat solution mathematically.

@dirklutz2818 5 ай бұрын

n³=(n+2)(n). n²=(n+2). n²-n-2=0. (n-2)(n+1)=0 Therefore n=2.

@KDev-fq9iq 5 ай бұрын

@@dirklutz2818smart answer 👍🏾

@Hexer1985 5 ай бұрын

In other words, you solved by trying... But actually, you need to prove, e.g. by induction, that there is no other solution in the domain of the natural numbers.

@sohaib_mer- 5 ай бұрын

then what is 0 !!

@vecenwilliams8172 5 ай бұрын

It's 1

@Syria_Free_Palestine_will_too 5 ай бұрын

@@vecenwilliams8172 and 2!!? also = 1?

@vecenwilliams8172 5 ай бұрын

@@Syria_Free_Palestine_will_too2!! is 2; it is just like 2!, which is 2*1 but you skip every other number (starting with the second) in this case just the "1" so it is just simple 2 (with no hidden ones). However 1!! is also 1.

@sardine_man 5 ай бұрын

Honestly from what I’ve seen it’s tricky, you can derive the answer 0!!=1 by just using pretty simple method of n!!=n*(n-2)!! And setting n=2 2!!=2*0!! 2!!/2=0!! 1=0!! But you can also arrive to other weird results by using the gamma function, 0!!=(2/pi)^0.5=sqrt(2/pi), honestly I think it’s hard to say with such radically different answers maybe one of them incorrect I’m not sure, but it’s either way not really defined since n!! Is usually defined for n € Z, n>0

@vecenwilliams8172 5 ай бұрын

@@sardine_man from what I have seen sources say that 0!! is 1 and like what you said it doesn't really make sense for it not to be an integer. I don't know the exact relationship with the gamma function and double factorials (only single factorials) so I would like to know what you did to get that answer? Also the gamma function was made (I think) to connect the relationship between whole numbers to get decimals, negatives, ect so the standard approach should be more correct, however you should still get an whole number if you plug in a whole number to the gamma function.

@JakeMarley-k6g 5 ай бұрын

I thought I was the only one who multipied by 5 that way

@jamesmarshall7756 5 ай бұрын

(n+2) divise n^3 donc il existe k appartenant à N tel que n^3=k(n+2). n=2 convient pour k=2 à l’évidence.

@nb_26xd 5 ай бұрын

( n + 2) = n * n * n

@alexandermorozov2248 5 ай бұрын

The usual search for options 🤣

@RyanLewis-Johnson-wq6xs 4 ай бұрын

n^3=(n+2)!! n ≈ -4.01543046331812587334352341617330739511771881380901757799609735840276126134131579513917295872962145862571141885425607963600894847385431053587846987720725846504946914362062967677547969294435656324390780730131492586337308371780708983830940247255278346585756033067702456202700425997450302757185645228011502418033153710863927768789265225708422435122931092975399736205276761084259839375095938524292624741520971880468008071384063829445798367946948304510...