I stumbled upon a specific series, who's Sum of squares of consecutive integers equals the sum of squares of the continuation of that consecutive integers.
For exmaple, this first number in the series results in: $3^2 + 4^2 = 5^2$.
The second results in: $10^2 + 11^2 + 12^2 = 13^2 + 14^2$.
The thrid: $21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2$. etc.
The formula for the series is:
By using; $1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}$ it is easy to prove that both side are the same.
Have I found something new? Useful?
I have a feeling that this is the only series with this specific property (consecutive sum of squares equal continuation of consecutive sum of squares), but I am unsure how to prove that.
Any comments would be appreciated.
$\endgroup$ 133 Answers
$\begingroup$$$\sum_{j=60}^{110} j^2 = \sum_{j=111}^{135} j^2$$ is a solution with $51$ terms on the left and $25$ terms on the right side.
So, the conjecture that there are always $x+1$ and $x$ terms, is false.
$\endgroup$ $\begingroup$Here are the examples of $\sum_{j=a}^{b} j^2 = \sum_{j=b+1}^{c} j^2$ where the limits are less than $1000$.
There are examples that do not follow your pattern. But I don't know whether there are other counterexamples beyond $1000$.
sum a b c L R 25 3 4 5 2 1 true 365 10 12 14 3 2 true 2030 21 24 27 4 3 true 7230 36 40 44 5 4 true 11900 18 34 42 17 8 false 19005 4 38 48 35 10 false 19855 55 60 65 6 5 true 42419 12 50 63 39 13 false 45955 78 84 90 7 6 true 94220 105 112 119 8 7 true 176460 136 144 152 9 8 true 308085 171 180 189 10 9 true 379525 60 110 135 51 25 false 508585 210 220 230 11 10 true 802010 253 264 275 12 11 true 963295 16 142 179 127 37 false 1217450 300 312 324 13 12 true 1254539 67 159 198 93 39 false 1789515 351 364 377 14 13 true 2558815 406 420 434 15 14 true 3572440 465 480 495 16 15 true 4884440 528 544 560 17 16 true 6556305 595 612 629 18 17 true 8657445 666 684 702 19 18 true
11265670 741 760 779 20 19 true
14467670 820 840 860 21 20 true
18359495 903 924 945 22 21 trueWell, $3^2+4^2=5^2$ has been around for quite a while; the rest is less ancient, but hardly new anyway.
Indeed, if we look up the OEIS for your sequence, the very first comment says:
Note that when starting from $a_n^2$, equality holds between series of first $n+1$ and next $n$ consecutive squares: $$a_n^2+(a_n+1)^2+\dots+(a_n+n)^2 = (a_n+n+1)^2+(a_n+n+2)^2+\dots(a_n+2n)^2;$$ e.g., $10^2+11^2+12^2 = 13^2+14^2$.
(I took the liberty of formatting the quote above. OEIS is run by seasoned math geeks who prefer to read their LaTeX code raw.)
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