## Poll

16 votes (50%) | |||

12 votes (37.5%) | |||

5 votes (15.62%) | |||

2 votes (6.25%) | |||

10 votes (31.25%) | |||

3 votes (9.37%) | |||

6 votes (18.75%) | |||

5 votes (15.62%) | |||

10 votes (31.25%) | |||

7 votes (21.87%) |

**32 members have voted**

Quote:teliotOkay ... new puzzle. A sum of consecutive squares equaling a sum of the following successive consecutive squares.

What's the longest one you can find?

Follow-up: as of now, it's 2,914,365 terms:

483,046^2 + ... + 2,699,107^2 = 2,699,108^2 + ... + 3,397,410^2 = 6,516,925,680,815,252,695

This is the 1254th such sequence (in order of the largest term to the left of the equals sign)

of course, the big question is if there are infinitely many solutions. If so, the first terms would be submissible to the encyclopedia of integer sequences.Quote:ThatDonGuyFollow-up: as of now, it's 2,914,365 terms:

483,046^2 + ... + 2,699,107^2 = 2,699,108^2 + ... + 3,397,410^2 = 6,516,925,680,815,252,695

This is the 1254th such sequence (in order of the largest term to the left of the equals sign)

Fabulous work by the way. I find this pretty cool, it says something about the density of sums of consecutive squares that there are so many solutions already.

Quote:teliotof course, the big question is if there are infinitely many solutions. If so, the first terms would be submissible to the encyclopedia of integer sequences.

I have an infinite family of solutions.

Let's use the notation introduced by ThatDonGuy:

Quote:ThatDonGuyLet there be A numbers on the left, and B on the right

(N - (A - 1))^2 + (N - (A - 2))^2 + ... + (N - 1)^2 + N^2 = (N + 1)^2 + (N + 2)^2 + ... + (N + B)^2

He derived the equation

Quote:ThatDonGuyN (N + 1) (2N + 1) - (N - A) (N - A + 1) (2 (N - A) + 1) = (N + B) (N + B + 1) (2 (N + B) + 1) - N (N + 1) (2N + 1)

This is a quadratic equation for N, the degree 3 terms cancel out. If you assume A=B+1, you can solve for N easily and get an integer solution.

Quote:teliotof course, the big question is if there are infinitely many solutions. If so, the first terms would be submissible to the encyclopedia of integer sequences.

Does the sequence have to be of known infinite length? The repunit primes is a sequence in OEIS, but I don't think it has been proven that it is of infinite length.

I don't know. Maybe curiosities will get in there if it's a well-known conjecture, but I wouldn't begin to submit unless I knew it was infinite and had no other obvious way of being generated.Quote:ThatDonGuyDoes the sequence have to be of known infinite length? The repunit primes is a sequence in OEIS, but I don't think it has been proven that it is of infinite length.

Quote:ThatDonGuyDoes the sequence have to be of known infinite length? The repunit primes is a sequence in OEIS, but I don't think it has been proven that it is of infinite length.

OEIS sequences do not need to be infinite. There are quite a number of finite sequences that have been accepted.

Is there a solution that starts at N = 1?Quote:ThatDonGuyFollow-up: as of now, it's 2,914,365 terms:

483,046^2 + ... + 2,699,107^2 = 2,699,108^2 + ... + 3,397,410^2 = 6,516,925,680,815,252,695

This is the 1254th such sequence (in order of the largest term to the left of the equals sign)

by the way, the smart folks here in this thread should come up with a new sequence, call it the wizard sequence, and submit it to OEIS.Quote:teliotIs there a solution that starts at N = 1?

Quote:teliotIs there a solution that starts at N = 1?

No - at least, none that I have found so far.

Pretty much every odd number from 3 through 99 has a sequence that has that many terms. I think the smallest even number size is 52.

Quote:ThatDonGuyPretty much every odd number from 3 through 99 has a sequence that has that many terms. I think the smallest even number size is 52.

My infinite family gives a solution for every odd number at least 3. The smallest even number I have found was also 52.

If you solve the quadratic equation for n in terms of A and B, the expression under the square root is 36A^2B^2-3((A-B)^4-(A-B)^2) = 36A^2B^2-3(d^4-d^2), where d=A-B.

This number has to be perfect square for n to be an integer. If N>0 then d>0, d=1 gives the infinite family in my earlier post, if d>1, then sqrt(36A^2B^2-3(d^4-d^2))<6AB, so if it is an integer, it is at most 6AB-1, this gives a bound on AB, AB≤(d^4-d^2)/4, which gives the bound B≤d(d-1)/2, this can help with computer searches.

The original question did not specify that you could only consider positive integers, (-k)^2+(-k+1)^2+...+(-1)^2+0^2=1^2+2^2+...+k^2 trivially, but there are also some non-trivial examples involving squares of some negative numbers, for example, (-2)^2+(-1)^2+0^2+1^2+...+7^2=8^2+9^2, or (-24)^2+...+14^2=15^2+...+27^2.