Tag: sums of squares

The familiar identity

$$3^2+4^2=5^2$$

can be viewed as an equality between two consecutive runs of squares. Surprisingly, this is the first member of a simple infinite family.

Pythagorean runs

This classical family is known as Pythagorean runs. For every positive integer $K$, consider

$$\sum _{j=0}^K{(A+j)}^2=\sum _{j=K+1}^{2K}{(A+j)}^2.$$

Thus the first $K+1$ consecutive squares have the same sum as the next $K$ consecutive squares.

There is exactly one positive value of $A$ for each $K$:

$$A=K(2K+1).$$

To see this, move the right-hand side to the left and simplify. The difference factors as

$$\sum _{j=0}^K{(A+j)}^2-\sum _{j=K+1}^{2K}{(A+j)}^2$$ $$=(A+K)(A-K(2K+1\left)\right).$$

Since $A>0$, the factor $A+K$ cannot vanish. Therefore

$$A=K(2K+1).$$

So there is a Pythagorean run for every $K$.

The first few are

$$3^2+4^2=5^2,$$ $${10}^2+{11}^2+{12}^2={13}^2+{14}^2,$$

and

$${21}^2+{22}^2+{23}^2+{24}^2$$ $$={25}^2+{26}^2+{27}^2.$$

This naturally raises another question:

What happens if squares are replaced by cubes?

Looking for cubic runs

The most direct analogue is to look for two runs of consecutive cubes having the same sum. In other words, this is the case $d=1$, where the terms on the right also differ by 1.

A computer search found no nontrivial examples. This does not prove that none exist, but it suggests looking at the next possibility.

For $d=2$, we compare $2m$ consecutive cubes with $m$ later cubes whose indices differ by 2:

$$\sum _{j=0}^{2m-1}{(A+j)}^3=\sum _{j=0}^{m-1}{(B+2j)}^3,$$ $$B>A+2m-1.$$

A computer search produced the remarkable example

$$\sum _{j=0}^{175}{(705+j)}^3=\sum _{j=0}^{87}{(913+2j)}^3.$$

Written out, this is

$${705}^3+{706}^3+\cdots +{880}^3$$ $$={913}^3+{915}^3+\cdots +{1087}^3.$$

Both sides equal

$$88681384000.$$

Here

$$(m,A,B)=(88,705,913).$$

At first, such a large numerical identity looks like something that might have occurred by accident. But it is actually the beginning of a much deeper pattern.

A Pell equation appears

Introduce the centered variables

$$X=2A+2m-1,$$ $$Y=B+m-1.$$

For the example above,

$$X=1585,Y=1000,$$

so

$$X:Y=317:200.$$

We therefore set

$$X=317z,Y=200z.$$

Substitution into the cubic-run equation and simplification lead to

$$48329z^2-156m^2=161.$$

This is a generalized Pell equation.

Our first cubic run corresponds to

$$(z,m)=(5,88),$$

since

$$48329{\left(5\right)}^2-156{\left(88\right)}^2=161.$$

The importance of the Pell equation is that one solution need not stand alone. Pell equations can generate further integer solutions, and in this case they produce infinitely many cubic runs.

Thus the huge identity found by computer search is not an isolated numerical curiosity. It belongs to an infinite Diophantine family.

The next cubic run

The next ordered solution is already enormous:

$$m=12005615252944088,$$

with

$$A=96105931914993680$$

and

$$B=124412740794926913.$$

Therefore the next run can be written as

$$S=\sum _{j=0}^{24011230505888175}{(96105931914993680+j)}^3,$$ $$S=\sum _{j=0}^{12005615252944087}{(124412740794926913+2j)}^3.$$

The final cube on the right has index

$$148423971300815087.$$

The enormous jump from $m=88$ to

$$m=12005615252944088$$

helps explain why these identities are so difficult to discover by direct search.

The computer found the first example.

The Pell equation explains why there are infinitely many more.

Reference

Michael Boardman, “Proof Without Words: Pythagorean Runs,” Mathematics Magazine, Vol. 73, No. 1 (2000), p. 59.

See also the corresponding sequence and additional historical references in OEIS A059255.