Published January 1999,February 2011.

One of the earlier NRICH articles Sums of Squares and Sums of Cubes has spurred me into further thought which I would like to share with the NRICH members. I would like to put forward some work I have done on the representation of numbers as sums of squares, and pose some further questions.

Consider some number $m$ such that: $$m = x^2 + y^2 + z^2.$$
The purpose of the following argument is to determine the form
which $m$ can take such that it can be written as the sum of three
squares. We shall begin by considering the situation where
$4^\alpha$ divides $m$ such that $\alpha$ is a positive integer. If
we consider $x$, $y$ and $z$ to be interchangeable then there are
four different selections of $x$, $y$ and $z$ being odd or even,
and these give the following situation. \begin{eqnarray} \mbox{3
odd numbers}: &x^2+y^2+z^2 \equiv 3 &\pmod 4\\ \mbox{2 odd
numbers}: &x^2+y^2+z^2 \equiv 2 &\pmod 4\\ \mbox{1 odd
number}: &x^2+y^2+z^2 \equiv 1 &\pmod 4\\ \mbox{0 odd
numbers}: &x^2+y^2+z^2 \equiv 0 &\pmod 4\\
\end{eqnarray}

Hence if 4 divides $m$, we have all of $x$, $y$ and $z$ even
and we can therefore divide $m$ by 4 and each of $x$, $y$ and $z$
by 2. We continue this procedure until not all of $x$, $y$ and $z$
are even.

The congruence $x^2\equiv-1\pmod {p}$, where $p$ is a prime,
has a solution $x$ if and only if $p\equiv 1\pmod {4}$ or
$p=2$.

If $p=2$ then we can see the solution $x=1$ and our work is
finished, so we assume that $p> 2$. Now take the congruence and
raise both sides to the power $(p-1)/2$, hence we have:
$$(-1)^{p-1\over2} \equiv \left(x^2\right)^{p-1\over2} \equiv
x^{p-1} \pmod p$$ However, we know by Fermat's Little Theorem that
$$x^{p-1}\equiv 1\pmod p \,\Rightarrow \, (-1)^{p-1\over2}\pmod p$$
However, since $p> 2$ we have $$-1 \neq 1\pmod p$$ which in turn
means that $(p-1)/2$ is even, or more exactly that $p$ is $1\pmod
4$, as required. So we have proved that the congruence has a
solution only if $p\equiv 1\pmod 4$. It is true, but harder to
prove, that it has a solution if $p\equiv 1\pmod 4$. I shan't do
that here.

The congruence $x^2\equiv-2\pmod p$, where $p$ is a prime, has
a solution $x$ if and only if $p=2$ or $p\equiv 1\pmod 8$ or
$p\equiv 3\pmod 8$.

We begin with a similar tack with our application of Fermat's
Little Theorem, which initially produces the following congruence
which needs to be solved: $$(-2)^{p-1\over2} \equiv
2^{p-1\over2}(-1)^{p-1\over2}\equiv 1\pmod p$$ We observe that
$p=2$ is a solution and now assume $p> 2$. The next step is to
show that the following identity is true: $$2^{p-1\over
2}\equiv(-1)^{p+1\over4}\pmod p$$ Consider the least positive
residues of the integers $2,4,6, . . . ,p-1$ when taken modulo $p$.
The number of these can be seen to be $(p-1)/2$, so the number
greater than $p/2$ is clearly $\left\lfloor{(p+1)/4}\right\rfloor$.
If we say $r_j$ are the residues greater than $p/2$, and $s_i$
those less than $p/2$, with $n$ being the number of $r_j$ and
$k=(p-1)/2-n$ being the number of $s_i$, then we wish to consider
the following product: $$(p-r_1)(p-r_2) \cdots (p-r_n)s_1s_2 \cdots
s_k$$ Now, the numbers $p-r_j$ and $s_j$ are all distinct. If they
were not then we would have: $$p-r_j=s_i \,\Rightarrow \, p=r_j+s_i
\,\Rightarrow \, 2|p$$ (2 would have to divide $p$ since all the
integers under consideration are even.) Since $p\neq 2$, this is
impossible; so we have a distinct set of integers. There are
$(p-1)/2$ of them, and all are less than $p/2$. So they must be
$1,2, . . . ,(p-1)/2$. Therefore: $$\eqalign{ (p-r_1)(p-r_2) \dots
(p-r_n)s_1s_2 \dots s_k &= 1\cdot2\dots{p-1\over2}\cr
(-r_1)(-r_2)\dots (-r_n)s_1s_2 \dots s_k &\equiv
1\cdot2\dots{p-1\over2} \pmod p\cr r_1r_2\dots r_ns_1s_2 \dots s_k
&\equiv (-1)^n\cdot1\cdot2\dots{p-1\over2} \pmod p\cr }$$
However, remember that the product of all the $r_j$ and $s_i$ is
simply the original product $2\cdot4\dots(p-1)$: $$2^{p-1\over2}
\cdot 1\cdot2\dots{p-1\over2} \equiv r_1r_2\dots r_ns_1s_2\dots s_k
\equiv (-1)^n \cdot 1\cdot2\dots{p-1\over2} \pmod p $$ so
$$(-1)^n\equiv 2^{p-1\over2} \pmod p$$ But earlier we noticed that
$n={\left\lfloor{p+1\over4}\right\rfloor}$; so, as required,
$2^{p-1\over 2}\equiv(-1)^{p+1\over4}\pmod p$. The congruence we
had to solve was $$2^{p-1\over2}(-1)^{p-1\over2} \equiv1\pmod p,$$
and we now see that this is equivalent to
$$(-1)^{\left\lfloor{p+1\over4}\right\rfloor}(-1)^{p-1\over2}
\equiv1\pmod p.$$ This means that we have to determine when
${\left\lfloor{(p+1)/4}\right\rfloor}+(p-1)/2$ is even, in the
knowledge that $p$ is odd. We consider the following to be possible
forms of $p$: $$p\equiv -3,-1,1,3 \pmod 8$$ Upon substitution we
see that the following forms of prime are the only solutions which
will work, and they will always have solutions: $$p\equiv \hbox{1
or 3}\pmod 8$$ This produces the required result for the Theorem.
(Again, what we have proved is that the congruence has a solution
only if $p\equiv \hbox{1
or 3}\pmod 8$. And again, it's possible, but harder to prove,
if .)

A prime $p$ can be represented in the form $a^2+b^2$ if the
congruence $x^2\equiv-1\pmod p$ is solvable.

Consider pairs of integers $(m,n)$ and let
$R=\left\lfloor\sqrt p\right\rfloor$. Hence we have $R<
\sqrt{p}p$ different pairs. Now look at the $r$ for which
$$m+nx\equiv r\pmod p$$ There are more pairs under consideration
than distinct residues, hence by the pigeonhole principle we have:
$$\eqalign{ &m_1 + n_1x \equiv m_2+n_2x \pmod p\cr \Rightarrow
\; &m_1-m_2 \equiv -x(n_1-n_2) \pmod p\cr \Rightarrow \;
&(m_1-m_2)^2 \equiv x^2(n_1-n_2)^2 \equiv -(n_1-n_2)^2 \pmod p
\quad\hbox{since $x^2\equiv 1$}\cr \Rightarrow \;
&(m_1-m_2)^2+(n_1-n_2)^2\equiv 0 \pmod p\cr }$$ But we know
from our earlier statements that $$(m_1-m_2)^2+(n_1-n_2)^2 \leq
2R^2 < 2p.$$ Also, since the pairs above are distinct these
squares cannot both be 0, hence we have $$0 <
(m_1-m_2)^2+(n_1-n_2)^2 < 2p$$ and so
$(m_1-m_2)^2+(n_1-n_2)^2=p$. This completes the theorem.

A prime $p$ can be represented in the form $a^2+2b^2$ if the
congruence $x^2\equiv-2\pmod p$ is solvable.

We employ an almost identical proof for this theorem as we did
for the last one. Consider pairs of integers $(m,n)$ and let
$R=\lfloor \sqrt p \rfloor$. Hence we have $R< \sqrt{p}p$
different pairs. Now look at the $r$ for which $$m+nx\equiv r\pmod
p$$ There are more pairs under consideration than distinct
residues, hence by the pigeonhole principle we have: $$\eqalign{
&m_1 + n_1x \equiv m_2+n_2x \pmod p\cr \Rightarrow \;
&m_1-m_2 \equiv -x(n_1-n_2) \pmod p\cr \Rightarrow \;
&(m_1-m_2)^2 \equiv x^2(n_1-n_2)^2 \equiv -2(n_1-n_2)^2 \pmod p
\quad\hbox{since $x^2\equiv -2$}\cr \Rightarrow \;
&(m_1-m_2)^2+2(n_1-n_2)^2\equiv 0 \pmod p\cr }$$ But we know
from our earlier statements that $$(m_1-m_2)^2+2(n_1-n_2)^2 \leq
3R^2 < 3p$$ Also, since the pairs above are distinct these
squares cannot both be 0, hence we have $$0 <
(m_1-m_2)^2+(n_1-n_2)^2 < 3p$$ and so $(m_1-m_2)^2+(n_1-n_2)^2$
is either $p$ or $2p$. Suppose $p$ is an odd prime such that
$a^2+b^2=2p$, then $a$ is obviously even. If $a$ is even then 4
divides $a^2$, so 4 cannot divide $2b^2$, so $b$ is odd. But then
we have $$2a^2+4b^2 = 4p \,\Rightarrow \, b^2 +
2\left({a\over2}\right)^2 = p.$$ Thus we conclude that any odd
prime satisfying the stated congruence can be expressed in the
stated form, which completes the theorem since clearly it's true
for $p=2$.

The product of any two numbers of the form $a^2+b^2$ can also
be written in that form.}

$$(a^2+b^2)(m^2+n^2) = (am-bn)^2 + (an+bm)^2$$ This algebraic
identity proves the Theorem.

The product of any two numbers of the form $a^2+2b^2$ can also
be written in that form.

$$(a^2+2b^2)(m^2+2n^2) = (am+2bn)^2 + 2(an-bm)^2$$ This
algebraic identity proves the Theorem.

Observe that 2 can be written as the sum of squares, and also
note the following identity: $$2(a^2+2b^2) = (2b)^2+2a^2$$ and it
can be expressed as $$2(a^2+b^2) = (a-b)^2 + (a+b)^2$$ wtf? The
purpose of establishing the above theorems is to try to put forward
a coherent theory on the sums of three squares. I would suggest
that all numbers except those of the form $4^\alpha(7+8k)$ can be
represented as the sum of three or less squares. Combining Theorems
2, 4 and 6 we observe that any number whose canonical factoring
takes the form $$2^\alpha\prod q^\beta\prod p^\gamma$$ where $q$
and $p$ are of the form 1 and 3 mod 8 respectively can be
represented as the sum of three squares. Similarly any number with
the canonical factoring of the form $$2^\alpha\prod q^\beta\prod
p^\gamma$$ where $q$ and $p$ are of the form 1 and 5 mod 8
respectively can be represented as the sum of three squares.
Obviously I have not shown that all numbers except those of the
form $4^\alpha(7+8k)$ can be represented as the sum of three
squares, but I would like to put that to anyone who is interested.
Also it would be interesting to know if the sum was unique or
otherwise. I hope that the above has been of interest and you will
be spurred to response!