A prime is in at most 1 way the sum of 2 squares

EWD1154 dealt with D.Zagier’s proof that a prime of the form 4k+1 is the sum of 2 squares. In fact, such a prime is in only 1 way the sum of 2 squares. In this note we show this by proving that if an odd n is the sum of 2 different pairs of squares, then that n is not prime.

Let an odd n be the sum of 2 squares; then the one square is odd, the other is even: the squares are of different parity. Let n be the sum of 2 squares in 2 ways; then there exist positive a, b, c, d such that

 (0) (a + b)² + (c – d)² = n (a – b)² + (c + d)² = n

(Here a is the average of the numbers of the one parity, c is the average of those of the other parity. Because we are considering distinct square decompositions, also b and d can be chosen positive.)

Eliminating n from (0) by equating the left-hand sides, we deduce after simplification

 (1) ab = cd       ,

from which we deduce the existence of positive r, s, t, v such that

 (2) a = sv b = rt c = st d = rv

(Consider “s := a gcd c;   v := a/s;   t := c/s;   r := b/t”.)

Now we observe
n
=         {(0)}
(a + b)² + (cd
=         {(1)}
a² + b² + c² + d²
=         {(2)}
s²v² + r²t² + s²t² + r²v²
=         {algebra}
(s² + r²) ∙ (t² + v²)

and because the 4 variables are positive, the two factors are each at least 2, and hence n is not a prime number.

*         *         *

The above was written down in Abilene State Park. In contrast to the proof discussed in EWD1154, I designed this proof myself, but the title of this note does not mention “derivation” of the proof, since I did not “derive” it in any technical sense.

I have considered investigation of the situation x²+y² = n u²+v² = n prime.n with the aim of showing (x,y) = (u,v) (x,y) = (v,u), but rejected that approach for the disjunction, and for the fact that I saw no way of using n’s primality. So I did some shunting and set myself to show that n was composite by writing it as a product of 2 plurals. I knew my complex numbers, in particular, that the modulus of a product is the product of the moduli, and then discovered that there was no point in looking at (x+yi)·(u+vi). Hence

 (3) (sv – rt)² + (st + rv)² = (s² + r²) · (t² + v²)

—the 2 expressions for the modulus of (s + ri) · (t + vi), which do equate a sum of squares to a product— has to be used differently. The right-hand side being even in r, it also equals (sv + rt + (strv)², and now we see the a, b, c, d entering the picture. The introduction of a ± b and c ± d circumvented the disjunctive complication of comparing unordered pairs.

I think I knew (3) outside the context of complex numbers as well; it is very common to separate in (a ± b)² the squares from the cross product, as in

 (a + b)² = (a – b)² + 4ab (a + b)² + (a – b)² = 2(a² + b²)       .

The proof reported provides a striking example of a proof in which the algebra is totally trivial while all subtlety has been invested in the decision what to name.

Austin, 7 June 1993

prof.dr. Edsger W.Dijkstra
Department of Computer Sciences
The University of Texas at Austin
Austin, TX 78712-1188
USA