# a hit I don't understand

Don Reble djr at nk.ca
Wed Apr 13 14:13:02 CEST 2005

```Wouter Meeussen observes that for
> Numbers [k] of the form 2^n or 5*2^n...
> k^2 cannot be a sum of three non-zero squares.
And also that each other positive square is a sum of three non-zero
squares.

Thanks, Wouter, for that interesting observation.

---

It's easy to prove the first part, that 4^n and 25*4^n aren't
representable (as the sum of three positive squares). Obviously, 1 and
25 aren't representable. Let X be the least representable number of the
form 4^n or 25*4^n.
Since X isn't 1 nor 25, X is a multiple of 4.
So X is either odd^2+odd^2+even^2 or even^2+even^2+even^2.
The first case is impossible, because odd^2 is 1 mod 4,
even^2 is 0 mod 4, and the sum is 2 mod 4; but X is 0 mod 4.
The second case is impossible, because then X/4 would be
representable, but X is the least.
So there's no least representable number, and therefore no representable
number, of that kind.

---

It's harder to prove the second part, that every other square is
representable. A difficult lemma:

For each prime P except 2 or 5, P^2 is representable.
In fact, the number of representations is
floor(P/8), if P == 1 mod 4;
ceil(P/8),  if P == 3 mod 4.
I started a thread in sci.math, and got a proof from Paul Pollack. (Thanks!)
Subject: "p^2 = a^2 + b^2 + c^2". (Watch out for the typos in the last
paragraph of LEMMA 2.) (And please ignore my dumb reply to Gerry. Thanks
to Paul, I know where I went wrong.)

Also, 25^2 = 20^2 + 12^2 + 9^2.

Given those, one can show that for any positive number X divisible by a
prime other than 2 or 5, or divisible by 25, X^2 is representable.
Let X = R*S, where R is one of those factors.
Then R^2 is representable as A^2 + B^2 + C^2,
and X^2 is representable as (S*A)^2 + (S*B)^2 + (S*C)^2.
Furthermore, that set of X's has every positive number except 2^n and 5*2^n.

--
Don Reble  djr at nk.ca

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