Sum of square patterns

Question

Can anyone give the name of this pattern

$$136^2+137^2+138^2+139^2+140^2+141^2+142^2+143^2+144^2 =\\ 145^2+146^2+147^2+148^2+149^2+150^2+151^2+152^2$$

Answer 1

It belongs to a family that starts with the familiar,

$$\begin{align} &3^2+4^2 = 5^2\\ &10^2+11^2+12^2=13^2+14^2\\ &21^2+22^2+23^2+24^2 = 25^2+26^2+27^2\\ \vdots\\ &a^2 + (a+1)^2 + \dots + a_n^2 = b^2 + (b+1)^2 + \dots + b_{n-1}^2 \end{align}$$

where $b = a_n+1$ and $a = 2n^2-3n+1 = 0, 3, 10, 21, 36, 55, 78, 105, \color{brown}{136},\dots$ Yours was the case of $n=9$ squares. In general, since,

$$a^2+(a+1)^2+\dots+(a+p-1)^2 = (p/6)(6a^2-6a + 6ap + 1 - 3p + 2p^2)$$

then you need to solve an equation of the form,

$$p(6a^2-6a + 6ap + 1 - 3p + 2p^2) = q(6b^2 - 6b + 6b q + 1 - 3q + 2q^2)\tag1$$

in positive integers. A parametric solution to $(1)$ is,

$$a = 2p^2-3p+1$$ $$b = a+p$$ $$q = p-1$$

for any $p$. The "continuous" case $b=a+p$ like the family above also has sporadic solutions like the $a,p,q=4,35,10$ given by bobbym, though these do not seem to belong to a second family.

Answer 2

This is actually more general: namely, define $s_n = \sum_{i=0}^{n} i = n(n+1)/2$. Then we check that $$\begin{split} s_n^2 + \dots + (s_n + n)^2 &= (s_n+n+1)^2 + \dots + (s_n + 2n)^2 \\ &= 4n^5 + 10n^4 + 25/3 n^3 + 5/2 n^2 + 1/6 n.\end{split}$$ (Both sides are necessarily polynomials of degree 5 as integrals of quartic polynomials, so it is enough to interpolate on the first 5 values). The example you use is just the case $n = 8$ of this formula.

Now there may be something even more conceptual (such as a trick involving geometric drawings, or even modular forms), but I don't get it right now.

Answer 3

If we consider, $3^2+4^2=5^2$ as triangle, $10^2+11^2+12^2=13^2+14^2$ as pentagon, $21^2+22^2+23^2+24^2=25^2+26^2+27^2$ as heptagon.

Then, interestingly, every structure has a right angled triangle as below.

Triangle: $3^2+4^2=5^2,$ Pentagon: $5^2+12^2=13^2,$ Heptagon: $7^2+24^2=25^2.$

The least length leg gives the count of the series as explained above in every right angled triangle.

Trigonometric Structure of the above series gives a common pattern that looks like a Olympic Torch or similar ones.

Answer 4

If we consider the equation of a certain type.

$$(n-4)^2+(n-3)^2+(n-2)^2+(n-1)^2+n^2+(n+1)^2+(n+2)^2+(n+3)^2+$$

$$+(n+4)^2=(k-3)^2+(k-2)^2+(k-1)^2+k^2+$$

$$+(k+1)^2+(k+2)^2+(k+3)^2+(k+4)^2$$

The solutions can be written simply using the sequence. The next element which is obtained from the previous one.

$$p_2=3p_1+4s_1$$

$$s_2=2p_1+3s_1$$

If we use the first element of the sequence. $( p ; s) - (2 ; 1)$

Then the formula will look like.

$$n=2p^2\pm6ps+4s^2$$

$$k=2p^2\pm6ps+5s^2$$

If we use the first element of the sequence. $(p ; s ) - ( 3 ; 2 )$

Then the formula will look like.

$$n=2(2p^2\pm6ps+4s^2)$$

$$k=2(2p^2\pm6ps+5s^2)$$