@@rls5907 Yes, you are the only person who took the time to characterize the problem as “contrived”. Would you share with us what you mean by contrived?
@MyOneFiftiethOfADollar I think that you describe exactly what I mean 60 seconds before the end of the video. At that time, you state that it’s not clear exactly what the purpose of this problem is. For me, it comes across as a fairly contrived problem (I.e. artificially created for the sake of it - rather than to show some interesting result). In the video you yourself say that many people will not find this problem interesting. To be clear, I enjoyed the video, watched it through to the end, but was surprised that I’m the only person to comment.
The notion of interesting is almost as vague as the notion of contrived. I wouldn’t know a priori if even a single triangular number could be expressed as the sum of the squares of consecutive odd integers. Would you? The video showed there is only one such instance of this occurring. The problem is from an older number theory book by sierpinski titled 250 problems and was of interest to many in the 70s. There was a slight mistake in the way the problem was stated which was corrected in the video.
There is nice free PDF version of 250 problems with solutions if you just search Sierpinski on the Web. It was problem 224 and you will see the error in problem wording after reading the solution. Are you a math teacher?
You mentioned to me that you were a little unsure of the claims you were making in this video. I think I can see where. At 3:13 you state that k has to be 4. I don't think you can simply assert that from 4(4n^2 + 1) = k(k + 1). At a first glance, why couldn't k be any multiple of 4? Or why couldn't k+1 be the term that is a multiple of 4. You need to probe a little deeper to eliminate those possibilities. Let's write S(n) for the sum of the squares of the consecutive odd integers (2n-1) and (2n+1). And write T(k) for the triangular number which is the sum of the integers from 1 to k. Then we have S(n) = 8n^2 + 2 and T(k) = k(k+1)/2, and we want them to be equal for some n, k ∈ ℕ. As you surmised, k = 4 looks likely, so we try it and get T(4) = 10 which is equal to S(1). Now we need to show that no other solutions exist when n > 1. So let's examine all the cases when k = 4n. Then we have T(4n) = 4n(4n+1)/2 = 8n^2 + 2n, which is always greater than S(n) = 8n^2 + 2 for n>1. But if we check k = (4n-1), we find T(4n-1) = (4n-1).4n/2 = 8n^2 - 2n, which is always less than S(n) = 8n^2 + 2. Since there is no integer between k-1 and k, there can be no value of k that makes T(k) = S(n) when n >1. Q.E.D. I hope that settles any unease you may have had about the proof you presented. Cheers!
@@RexxSchneider thx for writing back! k must be 4 since k(k+1) is the product of consecutive integers and is also = 4(4n^2 + 1) which cannot be equal to the product of consecutive integers unless n=1. I probably didn’t say or write enough during the video to convince viewers k has to be 4………I still have some ennui because 4(4n^2 + 1) logically does not have to be product of consecutive integers but think the uniqueness of the prime factorization/parity might play into claim k=4
Am I the only person who thought this was such a weirdly specific problem?! To the point of being contrived.
@@rls5907 Yes, you are the only person who took the time to characterize the problem as “contrived”.
Would you share with us what you mean by contrived?
@MyOneFiftiethOfADollar I think that you describe exactly what I mean 60 seconds before the end of the video. At that time, you state that it’s not clear exactly what the purpose of this problem is. For me, it comes across as a fairly contrived problem (I.e. artificially created for the sake of it - rather than to show some interesting result). In the video you yourself say that many people will not find this problem interesting.
To be clear, I enjoyed the video, watched it through to the end, but was surprised that I’m the only person to comment.
The notion of interesting is almost as vague as the notion of contrived. I wouldn’t know a priori if even a single triangular number could be expressed as the sum of the squares of consecutive odd integers. Would you?
The video showed there is only one such instance of this occurring.
The problem is from an older number theory book by sierpinski titled 250 problems and was of interest to many in the 70s. There was a slight mistake in the way the problem was stated which was corrected in the video.
I’ll check out the book. Thanks for the reference.
There is nice free PDF version of 250 problems with solutions if you just search Sierpinski on the Web. It was problem 224 and you will see the error in problem wording after reading the solution.
Are you a math teacher?
You mentioned to me that you were a little unsure of the claims you were making in this video. I think I can see where.
At 3:13 you state that k has to be 4. I don't think you can simply assert that from 4(4n^2 + 1) = k(k + 1). At a first glance, why couldn't k be any multiple of 4? Or why couldn't k+1 be the term that is a multiple of 4. You need to probe a little deeper to eliminate those possibilities.
Let's write S(n) for the sum of the squares of the consecutive odd integers (2n-1) and (2n+1). And write T(k) for the triangular number which is the sum of the integers from 1 to k.
Then we have S(n) = 8n^2 + 2 and T(k) = k(k+1)/2, and we want them to be equal for some n, k ∈ ℕ. As you surmised, k = 4 looks likely, so we try it and get T(4) = 10 which is equal to S(1). Now we need to show that no other solutions exist when n > 1.
So let's examine all the cases when k = 4n. Then we have T(4n) = 4n(4n+1)/2 = 8n^2 + 2n, which is always greater than S(n) = 8n^2 + 2 for n>1.
But if we check k = (4n-1), we find T(4n-1) = (4n-1).4n/2 = 8n^2 - 2n, which is always less than S(n) = 8n^2 + 2. Since there is no integer between k-1 and k, there can be no value of k that makes T(k) = S(n) when n >1. Q.E.D.
I hope that settles any unease you may have had about the proof you presented. Cheers!
@@RexxSchneider thx for writing back!
k must be 4 since k(k+1) is the product of consecutive integers and is also = 4(4n^2 + 1) which cannot be equal to the product of consecutive integers unless n=1. I probably didn’t say or write enough during the video to convince viewers k has to be 4………I still have some ennui because 4(4n^2 + 1) logically does not have to be product of consecutive integers but think the uniqueness of the prime factorization/parity might play into claim k=4