# CTK Insights

• ## Pages

11 Nov

### A Discovery of Hirotaka Ebisui And Thanos Kalogerakis

Today's communication from Thanos Kalogerakis brought to mind an insightful one page note by Alan Alda - a chapter in a collection This Explains Everything by John Brockman.

With every door into nature we nudge open, 100 new doors become visible, each with it own inscrutable combination lock.

On a rather small scale that Alan Alda's vision was paralleled in a message (and a short story that it conveyed) from Thanos Kalogerakis.

But again, as Alda puts it,

We need Einstein for GPS, but we can still get to the moon with Newton.

Who does not know or never heard of the Pythagorean theorem?

Early in the 19th century, a French geometer named Vecten studied a configuration of squares on the sides of a triangle. Vecten considered adding another layer of squares on the outer vertices of the Pythagorean squares, and then another one.

This configuration has many interesting properties, but Thanos Kalogerakis' message I mentioned at the outset, began with a statement, due to Hirotaka Ebisui. If we denote $A=a^2,$ $B=b^2,$ $C=c^2,$ then the Pythagorean theorem appears simply as $C=A+B.$

I shall denote the next layer of Vecten's squares by $A',B'C'$ and the next layer by $A'',B''C''$, as shown below:

Hirotaka Ebisui has found that in the case of the right triangle, $A'+B'=5C'$. On informing me of that result, Thanos added an identity for the next layer $A''+B''=C''.$

That was exciting enough for me to investigate. I can happily and proudly report that, for the next layer, $A'''+B'''=5C'''$.

Hence, I venture a general statement:

Let $A,B,C$ be the Pythagorean squares on the sides of a right triangle so that $A+B=C.$ Let's built recursively a sequence of triples of squares, $A_k,B_k,C_k$, with $A_0=A,$ $B_0=B,$ $C_0=C,$ and $A_{n+1},B_{n+1},C_{n+1}$ formed on the other vertices of $(B_n,C_n),$ $(C_n,A_n)$ and $(A_n,B_n).$ Then

$\begin{cases}A_n+B_n=C_n& n-even,\\A_n+B_n=5C_n& n-odd.\end{cases}$