Question

http://www.wolframalpha.com/input/?i=36%5E2%2B37%5E2%2B38%5E2%2B39%5E2%2B40%5E2%3D41%5E2%2B42%5E2%2B43%5E2%2B44%5E2&t=crmtb01

Answer 1

Can we always find a sequence like this for every odd number of consecutive numbers?

Answer 2

So for n=5 we have \(10^2+11^2+12^2=13^2+14^2\). My original post is for n=9 since there are 9 terms.

Answer 3

n=11 we have \(55^2+56^2+57^2+58^2+59^2+60^2=61^2+62^2+63^2+64^2+65^2\)

Answer 4

n=3 we have \(3^2+4^2=5^2\)

Haha sorry for these being in no particular order, just giving random examples. I'm just trying to find a proof that there will always be a sequence of an odd number of consecutive squares that satisfy this kind of relationship.

Answer 5

I have discovered a proof, but I'll let other people think about it if they like since this ended up having a pretty solution surprisingly!

Answer 6

http://www.wolframalpha.com/input/?i=%28%5Csum%5Climits_%7Bn%3D0%7D%5E6+%2878%2Bn%29%5E2%29+-+%28%5Csum%5Climits_%7Bn%3D7%7D%5E12+%2878%2Bn%29%5E2%29

Answer 7

i knew this pattern, il let others try too :)

Answer 8

I tried thinking of the simplest case and how to make it in a symmetric form that easily generalizes up. \[3^2+4^2=5^2\] So I rewrote this as \[(n-1)^2+n^2=(n+1)^2\] Now we will always add up in pairs to higher ones like this: \[(n-2)^2+(n-1)^2+n^2=(n+1)^2+(n+2)^2\] So what I did was subtract everything to the right hand side and turn it into a sum: \[n^2=\sum_{k=1}^a (n+k)^2-(n-k)^2\] This simplifies really nicely on the right hand side to: \[n^2=\sum_{k=1}^a 4nk=4n \sum_{k=1}^a k = 4n \frac{a(a+1)}{2}\] So we get that the middle most term is: \[n=2a(a+1)\] But that's kinda weird, we want the lowest term, so that we can count upwards, so we subtract \(a\) from this equation to get n to give us the first term. \[n=2a(a+1)-a = a(2a+1)\] This is super nice since not only does this form tell us the first term is \(n=a(2a+1)\) we also know that \(2a+1\) is the total number of terms altogether. If we were just to blindly write the sum out, we can just see that since we have \(a\) handy we know the middle term is \(a+1\) so we can just put the equals sign after that as we count off our fancy identity with little effort. =P

Answer 9

\(2a+1\) is total number of terms and n is the first term.

\(a*(2a+1)=n\)

\(1*(2*1+1)=3 \implies 3^2+4^2=5^2 \)
\( 2*(2*2+1)=10 \implies 10^2+11^2+12^2=13^2+14^2 \)
\( 3*(2*3+1)=21 \implies 21^2+22^2+23^2+24^2=25^2+26^2+27^2\)
etc... =)

Answer 10

you should check out OEIS https://oeis.org/
It's right up your alley

Answer 11

@FibonacciChick666 Yeah ever since I've seen @rational use it a handful of times I fell in love with that site haha.

Answer 12

lol, should have known

Answer 13

thats a pretty cool derivation! i bet you enjoyed deriving it as much i enjoyed reading it xD

Answer 14

Yeah it was fun and surprising. I wasn't expecting to be so lucky to find an answer this quickly. I'm glad to hear people enjoyed it!

Answer 15

easy peasy

Answer 16

:>

Answer 17

Hahaha so check it out, this generalizes further to we also have consecutive even, every third, etc... like this:

\(3^2+4^2=5^2\)
\(6^2+8^2=10^2\)
\(9^2+12^2=15^2\)

\(10^2+11^2+12^2=13^2+14^2\)
\(20^2+22^2+24^2=26^2+28^2\)
\(30^2+33^2+36^2=39^2+42^2\)

Fun and easy. But do the other pythagorean triples also generalize up in an interesting way as well to multiple terms?

Answer 18

so clever!