The Most Satisfying Proof: Product of 4 Consecutive Integers is not Square

Рет қаралды 24,622

Күн бұрын

Пікірлер: 58

@luckycandy4823 7 ай бұрын

Nice proposition, for the case of 4 consecutive integers by playing a bit with the expression we can say n(n+1)(n+2)(n+3)=n(n+3)(n+1)(n+2)=(n²+3n)(n²+3n+2)=(n²+3n+1-1)(n²+3n+1+1) And this is (n²+3n+1)²-1 by the known product formula (x-1)(x+1)=x²-1 So this is a bit more direct proof for this case.

@DrBarker 7 ай бұрын

This is very neat!

@teambellavsteamalice 7 ай бұрын

Indeed, I was going to post something similar, but thought I'd check the top comments first. I recently saw a puzzle in the form SQRT[ 1 + n(n+1)(n+2)(n+3) ] = some number (iirc 29).

@shinta69140 7 ай бұрын

But n(n+1)(n+2)(n+3)+1 IS Always a square number 😅

@williamperez-hernandez3968 7 ай бұрын

Two conclusions from this vid. 1. Adding 1 to the product of 4 consecutive integers is a square integer. 2. The requirement of positive integers is because if zero is one of the numbers, then the product is zero (which is a square). However if 1 is added to the product we get a perfect square independent if the four numbers are all negatives, or one of them is zero, in which case we get 1. So the requirement of positive integers is eliminated.

@JohnDavey-ve2su 7 ай бұрын

You also showed, in passing that n(n+1)(n+2)(n+3)+1 is always a square, which is very interesting also.

@davidbrisbane7206 7 ай бұрын

Actually, it is reasonably easy to show that n(n + 1)(n + 2)(n + 3) + 1 is a square. This being the case, then n(n + 1)(n + 2)(n + 3) can't be a square unless n = 0.

@Catman_321 7 ай бұрын

Another way to prove this in general for any number of consecutive integers (greater than or equal to 1) could go like this: All exponents in the prime factorization of a square number will be even. How could you manage to do this in general? Well, you just have to find a point where if you multiply enough consecutive integers, you will have an even exponent for all prime numbers we cover. Let's look at a base case of two consecutive integers: n(n+1). n and n+1 are trivially coprime which prevents them from sharing prime factors, so both n and n+1 would have to both be square. However, this never happens for n>0. What about 3 consecutive integers? Well, it gets a bit less obvious, but you can use a similar argument. Let's look at 2 different cases: n is either even or odd: if n is odd: all three numbers are coprime which doesn't work for the same reasons as stated in the case of 2 integers if n is even: n and n+2 share a factor of 2, so they must have a similar property that either n and n+2 are twice a square, and the other is half a square, and n+1 must be a square in the middle. This, again, won't happen for n>0. This game can continue (albeit it gets more complicated) for arbitrarily long strings which I won't continue here but that's the gist of it.

@armanavagyan1876 7 ай бұрын

Thanks DOCTOR BARKER.

@Gameboygenius 7 ай бұрын

Here's a silly idea I thought if when I saw that expression: Shifted by 1 for easier processing: (n-1)n(n+1)(n+2) Substitute: m=n+1 (n-1)(n+1)(m-1)(m+1)= (n²-1)(m²-1)=(n²-1)((n+1)²-1) One possibility this expression could be square is if both (n²-1) and ((n+1)²-1) are square, which they trivially aren't. The other possibility if you can prove something about the factors of (n²-1) and ((n+1)²-1) which might be a dead end.

@XxAspect23xX 7 ай бұрын

there's a much easier method we know in any 4 consecitive 2 numbers are going to be even and 1 of them is going to be divisible by 4 so the entire thing is going to be divisible by 8 and since the other 2 numbers are odd the largest power of 2 that would divide this product would be 8 or 2^3, thus we see that this is not possible as in it's prime factorisationn we wil have 2^3 which cannot be represented as a perfect square since (∑a^m b^n....)^2 is just a^2m + b^2n +c^2x....(where a,b,c oor simply ∑a are prime factors) thus for a number to be a perferct square all of it's prime factors must AND MUST have an even exponent as 2m is always even but here we have 2^3 and 3 is odd

@Deepnil 7 ай бұрын

This is asmr or smth 😂❤❤ love his videos.please bring complex analysis too

@mcwulf25 7 ай бұрын

Nice. I fiddled around with inequalities too.

@sr6424 7 ай бұрын

I tried solving the 4 consecutive in a different way. I set X= N + 1 and Y = X + 2. The expression comes to (X^2-1)(Y^2-1). A product of two numbers A and B equals ((A+B)/2)^2 - an integer. The difference between X^2 and Y^2 will always be negative hence putting these into the last formula will never produce an internet. Must thank you. Last night I put on a puzzle solving evening in a local bar! Inspired via various video content including yours.

@ingiford175 7 ай бұрын

You need to limit the scope. As there are solutions in integers to that. 0 * 1 * 2 *3 is a square. If you limit n to the naturals, you have a better case, but integers n in= {-3, -2, -2, 0} are four values taht give a square.

@ZipplyZane 7 ай бұрын

He specifically said that n is a positive integer (or natural number).

@pk2712 7 ай бұрын

These are very nice proofs .

@mcumer 7 ай бұрын

Another proof: n(n+1)(n+2)(n+3)=(n^2+3n)(n^2+3n+2)=(n^2+3n+1-1)(n^2+3n+1+1).. if we put a=n^2+3n+1, we can write tge product as (a+1)(a-1) which becomes a^2-1, that can't be a square

@eauna0029 7 ай бұрын

My first approach was based on divisibility, by seeing that n would need to divide (n+1)(n+2)(n+3) so it'd divide 6, and the case work would follow. And there probably is a generalisation in some way, but i was quite lazy to try and exploit divisibility properly for a more general proof.

@Nikioko 7 ай бұрын

n ⋅ (n + 1) ⋅ (n + 2) ⋅ (n + 3) = (n² + 3n) ⋅ (n² + 3n + 2) = (n² + 3n + 1 − 1) ⋅ (n² + 3n + 1 + 1) = (n² + 3n + 1)² − 1

@Bayerwaldler 7 ай бұрын

Erdös and Selfridge showed in 1975 that the product of two or more consecutive positive integers can never be a kth power for k>=2. So no need to go for a hunt…

@yuseifudo6075 7 ай бұрын

No there's a need

@OmarBlzk 7 ай бұрын

thank you sir.

@robertlunderwood 7 ай бұрын

I was thinking about a contradiction proof where it was a square and then using the Pythagorean Theorem to reduce it to a smaller square and then prove that the smaller square is in fact not a square.

@curtiswfranks 7 ай бұрын

How did you do those expansions so swiftly in your head?

@rkidy 7 ай бұрын

The magic of preparation

@MyOneFiftiethOfADollar 7 ай бұрын

What you have shown is very related to sqrt(n(n+1)(n+2)(n+3). + 1) = n^2 + 3n +1. Everybody and their brother has done this video e.g. sqrt(11(12)(13)(14)+1). = 11^2 + 33 +1 = 155

@romank.6813 7 ай бұрын

How about proving that n(n+1)(n+2)...(n+m) is never a square for any n and m?

@Anshusingh-p6z 7 ай бұрын

Why are u doing so multiply x,x+3 and x+1,x+2 and let x^2 + 3x = p then whole becomes p(p+2) which is obviously in between p^2 , (p+1)^2

@elmer6123 7 ай бұрын

(0+1+2+3)/4=3/2. Substitute n=m-3/2 to get (m-3/2)(m-1/2)(m+1/2)(m+3/2)=(m^2-1/4)(m^2-9/4)=m^4-(10/4)m^2+9/16. [m^4-(10/4)m^2+9/16]+[16/16]=(m^2-5/4)^2 is a perfect square for all m=n+3/2, so m^4-(10/4)m^2+9/16=n(n+1)(n+2)(n+3) cannot be a perfect square. I'm not a mathematician so there is probably a hole in this argument somewhere.

@grumpyparsnip 7 ай бұрын

The next obvious question is whether this works for any positive power of 2.

@guyhoghton399 2 ай бұрын

Let _x = n(n + 1)(n + 2)(n + 3)_ Let _m = n + ³/₂_ ∴ _x = (m - ³/₂)(m + ³/₂)(m - ½)(m + ½)_ = _(m² - ⁹/₄)(m² - ¼)_ Let _t = m² - ⁵/₄_ ∴ _x = (t - 1)(t + 1)_ ⇒ _x = t² - 1_ Since _x ∈ ℕ_ then _t ∊ ℤ_ Suppose that for some integer _a:_ _x = a²_ ∴ _t² - a² = 1_ ⇒ _(t - a)(t + a) = 1_ ⇒ _t - a = t + a = ±1_ ⇒ _a = 0_ ⇒ _x ∉ ℕ_ ⇒ *_x ≠ a²_*

@picrust314 7 ай бұрын

Would polynomial squarerooting be a valid proof since it does not end up as a integer expression?

@borekking4769 7 ай бұрын

Can this be generalized to any power or two?

@bjornfeuerbacher5514 7 ай бұрын

See the comment by Bayerwaldler: "Erdös and Selfridge showed in 1975 that the product of two or more consecutive positive integers can never be a kth power for k>=2"

@DrBarker 7 ай бұрын

I thought it was interesting that the argument worked for 2, 4, and 8, so I've just had a go with 16 consecutive positive integers. (n^8 + 60n^7 + 1490n^6 + 19800n^5 + 151761n^4 + 671580n^3 + 1609180n^2 + 1741200n + 430016 )^2 - n(n+1)...(n+15) is eventually positive (for large enough n), since the difference is a polynomial of degree six, with a positive coefficient of n^6. (n^8 + 60n^7 + 1490n^6 + 19800n^5 + 151761n^4 + 671580n^3 + 1609180n^2 + 1741200n + 430015 )^2 - n(n+1)...(n+15) is eventually negative (for large enough n), since the difference is a polynomial of degree eight, with a negative coefficient of n^8. This means that, for large enough values of n, n(n+1)...(n+15) lies in between the consecutive square numbers (n^8 + 60n^7 + 1490n^6 + 19800n^5 + 151761n^4 + 671580n^3 + 1609180n^2 + 1741200n + 430015 )^2 and (n^8 + 60n^7 + 1490n^6 + 19800n^5 + 151761n^4 + 671580n^3 + 1609180n^2 + 1741200n + 430016 )^2. So we can essentially do the same for 16 consecutive positive integers, but there might be a lot of individual cases to check for small values of n. Not sure if/how we could prove that this method will always work for the product of 2^k consecutive positive integers, but it would be really interesting to find out if this also works for 32 consecutive integers, or in general for all powers of 2.

@wychan7574 7 ай бұрын

@@DrBarkerFor all positive integers n, n(n+1)(n+2) can never be a square.

@wychan7574 7 ай бұрын

Proof three consecutive positive integers cannot be a square. Suppose n(n+1)(n+2) is a square for some positive n. consider two cases 1.(n+1) is a square. Then n(n+2)must also be a square. But n(n+2)=n^2+2n

@brickie9816 7 ай бұрын

Can this be somehow proven using coprime numbers? This was my first instinct but it didn't lead anywhere so now im just curious.

@RoyceDima 7 ай бұрын

Is there some other k such that there is a product of k consecutive integers that is square?

@Bayerwaldler 7 ай бұрын

No, there is no such k. Not even if you replace square by cube, fourth power and so on. This is a result of Erdös and Selfridge from 1975.

@PunkSage 5 ай бұрын

@@Bayerwaldler Interesting. I was also surprised by a dedication to W.Sierpiński at the begging of the paper: www.renyi.hu/~p_erdos/1975-46.pdf

@arifyesehehehehhewahahahah3445 7 ай бұрын

f(n) = n(n+1)(n+2)(n+3) f(0) = 0 √0 = 0 0 is an integer 0 is a square f(n) can be a square.

@martinflygar7699 7 ай бұрын

Are there any known examples where the product of consecutive numbers do make a square?

@KSignalEingang 7 ай бұрын

I've been wondering the same. I came up with a fairly simple inductive proof that no factorial can be square (except the trivial cases of 1! and 0!), and I feel like there's something to be made out of the fact that any product of k consecutive integers can be expressed as n!/(n-k)!, but I haven't managed to hammer out a proof. Nor have I found any counterexamples doing brute-force searches, though obviously that's no guarantee of anything (I've only done the search for k=3, and didn't test for particularly high values of n). (Edit: @Bayerwaldler answered this elsewhere in the comments, apparently the answer is a definitive no.)

@Happy_Abe 7 ай бұрын

Does this generalize to all powers of 2?

@DrBarker 7 ай бұрын

We can get the same argument to work for the product of 16 consecutive integers, with some small values of n to deal with separately - I added some more details to another comment here. It would be really interesting to know if this works for all powers of 2 though.

@Happy_Abe 7 ай бұрын

@@DrBarker thanks I saw another comment saying something about it being true for any consecutive product of k integers. If that’s true then would include all powers of 2 automatically but showing that must be much harder

@wuchinren 7 ай бұрын

hint: n(n+1)(n+2)(n+3)+1 must be a square.

@krwada 7 ай бұрын

Interesting ... All your coefficients appear to be divisible by 'm', where 'm' is the number of consecutive integers So, instead of brute-force, as you show in this video; Is it possible to show? 1. All the coefficients generated by a generic proof 2. A generalized proof where we have 'm' consecutive integers I would like to see a generalized proof where we have the product 'm' consecutive integers as 'm' approaches infinity never being equal to a square. PRODUCT = n(n+1)(n+2)(n+3) ... (n+m)