"Brandt Brauer Frick"

Written and directed by Danae Diaz and Patricia Luna
Music by Brandt Brauer Frick

"Caffeine" taken from the album "You Make Me Real"

(Used at Prix de Lausanne 2012 as a set piece of contemporary dance.)

Igor Fyodorovitch Stravinsky [1882-1971] wrote this requiem in his last years.

It is said that he wasn’t pursuing excellence of linkage of tone row

but a sense like “Polystylism” could be obtained from crossing between several styles.

His spirit of inquiry was immeasurable.

Listening this requiem, you will probably feel

something modern or misterious perhaps a little weird.

I guess it was calculated to the smallest detail by him.

The pencils which I had been longing to get came to hand.

How beautiful they are… I like them so much.

(The slim one’s product name is “slight” and the other one’s name is “compact”,both of them are from the stationery brand “woerther” established in Germany. )

Basic academic research in our school.

I’ve studied Elementary number theory,particularly Pell’s equation.

One problem I made last year puzzled me for a long time,

so I wanted to solve it on this occasion. The problem is…

For any natural number n,

show that there are countless natural numbers (a,b,c)

such that a^2+b^2=c^2 ∧ |a-b|=n.


Lemma: There are countless natural numbers (a,b,c) such that a^2+b^2=c^2 ∧ |a-b|=1

proof: No generality is lost by supposing that a≦b,then b=a+1,

thus a^2+b^2=c^2∧|a-b|=1



Suppose that 2a+1=X, c=Y, we obtain X^2-2Y^2=-1 …(1)

Now,we apply Brahmagupta’s Identity  (x^2+y^2)(z^2+w^2)=(xz−yw)^2+(xw+yz)^2

Let x=X,y=√2Yi,z=3,w=2√2i,

we get X^2−2Y^2=(3X+4Y)^2 +(2√2Xi+3√2Yi)^2=(3X +4Y)^2−2(2X +3Y)^2 …(2)

Since (X,Y)=(1,1) satisfies (1), furthermore (2) holds,

All natural number (X_n,Y_n) such that 

X_1=1,Y_1=1 ∧ X_(n+1)=3X_n+4Y_n ∧ Y_(n+1)=2X_n+3Y_n satisfies (1).

And obviously X_(n+1)>X_n, every (X_n,Y_n) is different from each other.

Furthermore, X^2=2Y^2-1 implies X is odd number,

thus a is always replaced with X by one-to-one correspondence because X=2a+1.

Therefore, there are countless natural numbers (a,b,c) 

such that a^2+b^2=c^2 ∧ |a-b|=1…(A)

Now, if we multiply (a,b,c) such that (A) by n,the product is (na,nb,nc).

For these (na,nb,nc),

a^2+b^2=c^2∧|a-b|=1 ⇔ (na)^2+(nb)^2=(nc)^2∧|na-nb|=n

thus For any natural number n, there are countless natural numbers (a,b,c) 

such that a^2+b^2=c^2 ∧ |a-b|=n.