# Sums of descending squares

I am interested in integers that can be expressed as a sum of squares.

Specifically I am interested in integers that can be expressed as follows:

$n=6*Sum (k^2+(k-a)^2+(k-2a)^2.....1^2)$

These values $N$ are congruent numbers* (not necessarily square free but if $n$ has a square it can be divided into $k$ and $a$ to make a square free if $k$ and $a$ are rationals).

*Congruent Numbers are solutions to the congruent number problem and are defined as

$N=X Y/2$

Where

$X^2+Y^2=Z^2$

Squarefree congruent numbers are where $X$, $Y$, and $Z$ can be rationals but $N$ has only Integer primes to the first power as factors.

Are there any tricks for finding a sequence of unique squares that sum to an integer like this?

The triplet that defines the $N$ value is

$X=2k+a$
$Y=2k+2k^2/a$
$Z=a+2k+(2k^2/a)$

I am interested in any tricks that turn an integer into a sum of squares other than brute force.

Thanks for your help!

PS I seem to remember if an integer is the sum of $2$ squares then those two squares are themselves integers does this hold for longer sequences at all and if so when?

**

Congruent numbers are typically of the form #(#+1)(2#+1) or #(#+1)(#+2)

But the sum of squares is interesting because congruent numbers can also split to be the sum of two or more sequences.

So

$n=6*Sum (k^2+(k-a)^2+(k-2a)^2.....(k-ma)^2 +(k-ma-b)^2 +(k-ma-2b)^2 . ..+...+..1^2)$

Where

$b= 2(k-ma)^2/( a)$ and $b<a$

• You are interested in integers of the form n(n+1)(2n+1)? – Vishaal Selvaraj Mar 7 '18 at 7:06