Happy St Patrick’s Day and Sum of Squares
A short proof of the sum of two square theorem

Neil L. is not a computer scientist–he is a Leprechaun. Since this Tuesday is St. Patrick’s Day it seems right to talk about him. As long as you look at him he cannot escape, but the moment you look away, he vanishes. I thought in the same spirit I would share a wonderfully simple proof that I read recently, it is due to Zaiger. The proof is based on a trick that reminded me a bit of Neil: when you read the proof carefully it clearly works, but if you look away it seems magical.
Theorem 1 Every prime
is the sum of two squares.
There are many proofs of this theorem, but the following is definitely very cool. It is based on a simple, but powerful parity principle.
Proof: Let be a prime. Consider the finite set
where
are natural numbers. This set has two involutions: the obvious
that sends
to
and the following:
that sends
to:
-
, if
;
-
, if
;
-
, if
.
Any fixed point of is of the form
and yields a representation of
as a sum of two squares:
The mapping has one fixed point
. This shows that
has odd cardinality; therefore,
must have a fixed point.
Magic.
Open Problems
Have a happy St. Patrick’s Day.


Thank you for sharing this. I notice you did not call it a “proof from the Book”, which is what I would have first thought of.
Do you really think its from the “book”? I not sure myself. I think it is definitely a very neat proof.
Thanks for this post! I had a great time checking all the claims. Now I just need to go read Heath-Brown’s paper, the only reference of Zaiger’s, and see if I can understand where the involution, g, came from.
Love you “hat”? Glad you enjoyed the post….
rjlipton
Actually Chapter 4 of “Proofs from THE BOOK” discusses precisely this problem. They present Heath-Brown’s proof, of which Zagier’s is a condensed version.
I was a bit suspicious of this proof that it never mentioned the need for p to be prime. This is hidden in the statement that g has only 1 fixed point. Nice proof!