Post History
#13: Post edited
by
beloh
·
2022-12-22T10:51:07Z (over 1 year ago)
Find all integer solution for $a_1(a_1+k)(a_1+2k)...(a_1+(2k+1)d)+1 = x^2$
- Find all integer solutions for $a_1(a_1+k)(a_1+2k)...(a_1+(2k+1)d)+1 = n^2$
- Find all integer solution for the equation below:
$$a_1(a_1+k)(a_1+2k)...(a_1+(2k+1)d)+1 = x^2$$To write it in iterated operator notation: $$ 1+\prod_{k=0}^{2k+1} (a_1 + kd) = x^2 $$
- Find all integer solution for the equation below:
- $$a_1(a_1+k)(a_1+2k)...(a_1+(2k+1)d)+1 = n^2$$
- =$$ 1+\prod_{k=0}^{2k+1} (a_1 + kd) = n^2 $$
#12: Question closed
by
beloh
·
2022-12-22T10:47:14Z (over 1 year ago)
#11: Post edited
by
Peter Taylor
·
2022-12-01T08:20:32Z (over 1 year ago)
When I fixed the consistency in the body earlier I overlooked the title
Find all integer solution for $a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = x^2$
- Find all integer solution for $a_1(a_1+k)(a_1+2k)...(a_1+(2k+1)d)+1 = x^2$
#10: Post edited
by
celtschk
·
2022-12-01T08:20:05Z (over 1 year ago)
Changed displaymath to inline math in title
Find all integer solution for $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = x^2$$
- Find all integer solution for $a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = x^2$
Find all integer solution for the equation below:
$$a_1(a_1+k)(a_1+2k)...(a_1+(2k+1)d)+1 = x^2$$
To write it in iterated operator notation: $$ 1+\prod_{k=0}^{2k+1} (a_1 + kd) = x^2 $$
#9: Post edited
by
Peter Taylor
·
2022-11-28T08:24:28Z (over 1 year ago)
Make the two expressions consistent with each other, use more standard limits
- Find all integer solution for the equation below:
$$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = x^2$$To write it in summation notation: $$ 1+\prod_{-1<k<2k+2} (a_1 + kd) $$
- Find all integer solution for the equation below:
- $$a_1(a_1+k)(a_1+2k)...(a_1+(2k+1)d)+1 = x^2$$
- To write it in iterated operator notation: $$ 1+\prod_{k=0}^{2k+1} (a_1 + kd) = x^2 $$
#8: Post edited
by
beloh
·
2022-11-24T13:41:46Z (over 1 year ago)
- Find all integer solution for the equation below:
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = x^2$$
$$ = 1+\prod_{-1<k<2k+2} (a_1 + kd) $$
- Find all integer solution for the equation below:
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = x^2$$
- To write it in summation notation: $$ 1+\prod_{-1<k<2k+2} (a_1 + kd) $$
#7: Post edited
by
beloh
·
2022-11-24T13:34:05Z (over 1 year ago)
Prove That $$ 1+\prod_{1<k<=2k+1} (a_1 + kd) $$ Is Going To Be A Square
- Find all integer solution for $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = x^2$$
This question raised when noticing the pattern while i was solving:$$\sqrt{500(501)(502)(503)+1}$$More formally in general terms:Given $ n \in \mathbb{N}^{\ast} ,$ Prove that the following is a square$$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = $$$$ 1+\prod_{1<k<=2k+1} (a_1 + kd) $$and when simplifying this polynomial you first try to see if its a perfect square or not so that you can cancel the square root...You might recognize this product as it being an [arithmetic progression](https://en.wikipedia.org/wiki/Arithmetic_progression#Product)(PS: It seems to be working with only value 4 other than that range also its the same with negative values as well)As a bonus see if you can extend it and go beyond natural numbers.
- Find all integer solution for the equation below:
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = x^2$$
- $$ = 1+\prod_{-1<k<2k+2} (a_1 + kd) $$
#6: Post edited
by
beloh
·
2022-11-19T20:25:43Z (over 1 year ago)
Prove That $$ 1+\prod_{k>=12} (a_1 + kd) $$ Is Going To Be A Square
- Prove That $$ 1+\prod_{1<k<=2k+1} (a_1 + kd) $$ Is Going To Be A Square
- This question raised when noticing the pattern while i was solving:
- $$\sqrt{500(501)(502)(503)+1}$$
- More formally in general terms:
- Given $ n \in \mathbb{N}^{\ast} ,$ Prove that the following is a square
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = $$
$$ 1+\prod_{0<=k<=2k+1} (a_1 + kd) $$- and when simplifying this polynomial you first try to see if its a perfect square or not so that you can cancel the square root...
- You might recognize this product as it being an [arithmetic progression](https://en.wikipedia.org/wiki/Arithmetic_progression#Product)
- (PS: It seems to be working with only value 4 other than that range also its the same with negative values as well)
- As a bonus see if you can extend it and go beyond natural numbers.
- This question raised when noticing the pattern while i was solving:
- $$\sqrt{500(501)(502)(503)+1}$$
- More formally in general terms:
- Given $ n \in \mathbb{N}^{\ast} ,$ Prove that the following is a square
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = $$
- $$ 1+\prod_{1<k<=2k+1} (a_1 + kd) $$
- and when simplifying this polynomial you first try to see if its a perfect square or not so that you can cancel the square root...
- You might recognize this product as it being an [arithmetic progression](https://en.wikipedia.org/wiki/Arithmetic_progression#Product)
- (PS: It seems to be working with only value 4 other than that range also its the same with negative values as well)
- As a bonus see if you can extend it and go beyond natural numbers.
#5: Post edited
by
beloh
·
2022-11-19T15:57:31Z (over 1 year ago)
- This question raised when noticing the pattern while i was solving:
- $$\sqrt{500(501)(502)(503)+1}$$
- More formally in general terms:
- Given $ n \in \mathbb{N}^{\ast} ,$ Prove that the following is a square
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = $$
$$ 1+\prod_{k>=12} (a_1 + kd) $$- and when simplifying this polynomial you first try to see if its a perfect square or not so that you can cancel the square root...
- You might recognize this product as it being an [arithmetic progression](https://en.wikipedia.org/wiki/Arithmetic_progression#Product)
- (PS: It seems to be working with only value 4 other than that range also its the same with negative values as well)
- As a bonus see if you can extend it and go beyond natural numbers.
- This question raised when noticing the pattern while i was solving:
- $$\sqrt{500(501)(502)(503)+1}$$
- More formally in general terms:
- Given $ n \in \mathbb{N}^{\ast} ,$ Prove that the following is a square
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = $$
- $$ 1+\prod_{0<=k<=2k+1} (a_1 + kd) $$
- and when simplifying this polynomial you first try to see if its a perfect square or not so that you can cancel the square root...
- You might recognize this product as it being an [arithmetic progression](https://en.wikipedia.org/wiki/Arithmetic_progression#Product)
- (PS: It seems to be working with only value 4 other than that range also its the same with negative values as well)
- As a bonus see if you can extend it and go beyond natural numbers.
#4: Post edited
by
beloh
·
2022-11-18T18:53:10Z (over 1 year ago)
Prove That $$ 1+\prod_{k=0}^{2n+1} (a_1 + kd) $$ Is Going To Be A Square
- Prove That $$ 1+\prod_{k>=12} (a_1 + kd) $$ Is Going To Be A Square
- This question raised when noticing the pattern while i was solving:
- $$\sqrt{500(501)(502)(503)+1}$$
- More formally in general terms:
- Given $ n \in \mathbb{N}^{\ast} ,$ Prove that the following is a square
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = $$
$$ 1+\prod_{k=0}^{2n+1} (a_1 + kd) $$- and when simplifying this polynomial you first try to see if its a perfect square or not so that you can cancel the square root...
- You might recognize this product as it being an [arithmetic progression](https://en.wikipedia.org/wiki/Arithmetic_progression#Product)
(PS: the variable $d$ is kind of redundant but i put it there to be consistent with the general form)- As a bonus see if you can extend it and go beyond natural numbers.
- This question raised when noticing the pattern while i was solving:
- $$\sqrt{500(501)(502)(503)+1}$$
- More formally in general terms:
- Given $ n \in \mathbb{N}^{\ast} ,$ Prove that the following is a square
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = $$
- $$ 1+\prod_{k>=12} (a_1 + kd) $$
- and when simplifying this polynomial you first try to see if its a perfect square or not so that you can cancel the square root...
- You might recognize this product as it being an [arithmetic progression](https://en.wikipedia.org/wiki/Arithmetic_progression#Product)
- (PS: It seems to be working with only value 4 other than that range also its the same with negative values as well)
- As a bonus see if you can extend it and go beyond natural numbers.
#3: Post edited
by
beloh
·
2022-11-18T09:33:03Z (over 1 year ago)
This question arisen when i was trying to solve:- $$\sqrt{500(501)(502)(503)+1}$$
- More formally in general terms:
- Given $ n \in \mathbb{N}^{\ast} ,$ Prove that the following is a square
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = $$
- $$ 1+\prod_{k=0}^{2n+1} (a_1 + kd) $$
and when simplifying this polynomial you first try to see if its a perfect square or not so that you can factor the square root...- You might recognize this product as it being an [arithmetic progression](https://en.wikipedia.org/wiki/Arithmetic_progression#Product)
- (PS: the variable $d$ is kind of redundant but i put it there to be consistent with the general form)
- As a bonus see if you can extend it and go beyond natural numbers.
- This question raised when noticing the pattern while i was solving:
- $$\sqrt{500(501)(502)(503)+1}$$
- More formally in general terms:
- Given $ n \in \mathbb{N}^{\ast} ,$ Prove that the following is a square
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = $$
- $$ 1+\prod_{k=0}^{2n+1} (a_1 + kd) $$
- and when simplifying this polynomial you first try to see if its a perfect square or not so that you can cancel the square root...
- You might recognize this product as it being an [arithmetic progression](https://en.wikipedia.org/wiki/Arithmetic_progression#Product)
- (PS: the variable $d$ is kind of redundant but i put it there to be consistent with the general form)
- As a bonus see if you can extend it and go beyond natural numbers.
#2: Post edited
by
beloh
·
2022-11-18T09:02:28Z (over 1 year ago)
Prove That x*(x+1)*(x+2)...(x+2n+1) Is Going To Be A Square
- Prove That $$ 1+\prod_{k=0}^{2n+1} (a_1 + kd) $$ Is Going To Be A Square
The question arisen when i was trying to find the solution to:- $$\sqrt{500(501)(502)(503)+1}$$
and when simplifying this polynomial you first try to see if its a perfect square or not so that you can factor the square root.Can you Prove that this will be true, for positive integers or beyond.
- This question arisen when i was trying to solve:
- $$\sqrt{500(501)(502)(503)+1}$$
- More formally in general terms:
- Given $ n \in \mathbb{N}^{\ast} ,$ Prove that the following is a square
- $$a_1(a_1+1)(a_1+2)...(a_1+(2k+1)d)+1 = $$
- $$ 1+\prod_{k=0}^{2n+1} (a_1 + kd) $$
- and when simplifying this polynomial you first try to see if its a perfect square or not so that you can factor the square root...
- You might recognize this product as it being an [arithmetic progression](https://en.wikipedia.org/wiki/Arithmetic_progression#Product)
- (PS: the variable $d$ is kind of redundant but i put it there to be consistent with the general form)
- As a bonus see if you can extend it and go beyond natural numbers.
#1: Initial revision
by
beloh
·
2022-11-17T21:58:03Z (over 1 year ago)
Prove That x*(x+1)*(x+2)...(x+2n+1) Is Going To Be A Square
The question arisen when i was trying to find the solution to:
$$\sqrt{500(501)(502)(503)+1}$$
and when simplifying this polynomial you first try to see if its a perfect square or not so that you can factor the square root.
Can you Prove that this will be true, for positive integers or beyond.