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.

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

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

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

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

Your a great techar well done man

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

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

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

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 👍

Cool problem and good analysis/solution.

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

New knowledge, sir!🎉🎉🎉

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.

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

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

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.

Nice analogy

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

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).

just love your ividios

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.

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.

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.

Thanks Sir

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.

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?

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

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.

Got it. Easy problem

What did you do after highschool?

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.

Can solve by Gama function ?

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*

(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.

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.

n=2 by inspection.

The usual search for options 🤣

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

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

Well then what’s 3!!!!

First

What if my neighbor is the enemy?

2.

then what is 0 !!

( n + 2) = n * n * n

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.