Lagrange's Four Square Theorem/Proof 1

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Theorem

Every positive integer can be expressed as a sum of four squares.


Proof

$1$ can trivially be expressed as a sum of four squares:

$1 = 1^2 + 0^2 + 0^2 + 0^2$


From Product of Sums of Four Squares it is sufficient to show that each prime can be expressed as a sum of four squares.


The prime number $2$ certainly can: $2 = 1^2 + 1^2 + 0^2 + 0^2$.


It remains to consider the odd primes.


Existence of $m: 1 \le m < p$ such that $m p$ is the sum of $4$ squares

Suppose that some multiple $m p$ of the odd prime $p$ can be expressed as:

$m p = a^2 + b^2 + c^2 + d^2, 1 \le m < p$

If $m = 1$, we have the required expression.

If not, then after some algebra we can descend to a smaller multiple of $p$ which is also the sum of four squares:

$m_1 p = a_1^2 + b_1^2 + c_1^2 + d_1^2, 1 \le m_1 < m$


Next we need to show that there really is a multiple of $p$ which is a sum of four squares.

From this multiple we can descend in a finite number of steps to $p$ being a sum of four squares.


Since $p$ is odd and greater than $2$, $\dfrac {p - 1} 2$ is an integer.

There are $\dfrac {p + 1} 2$ integers $a_1, a_2, \ldots$ such that $0 \le a_i \le \dfrac {p - 1} 2$.

For each $a_i$, let $r_i$ be the remainder when ${a_i}^2$ is divided by $p$.

We have:

$\forall r_i: 0 \le r_i \le p - 1$


Aiming for a contradiction, suppose:

$\exists a_1, a_2: 0 \le a_2 < a_1 \le \dfrac {p - 1} 2: {a_1}^2 = q_1 p + r, {a_2}^2 = q_2 p + r$

That is, two different integers in that range which have the same remainder.

Then:

\(\ds {a_1}^2 - {a_2}^2\) \(=\) \(\ds \paren {q_1 - q_2} p\)
\(\ds \leadsto \ \ \) \(\ds p\) \(\divides\) \(\ds \paren { {a_1}^2 - {a_2}^2}\)
\(\ds \) \(=\) \(\ds \paren {a_1 - a_2} \paren {a_1 + a_2}\)

By Euclid's Lemma for Prime Divisors, either:

$p \divides a_1 - a_2$

or:

$p \divides a_1 + a_2$

But both $a_1 - a_2$ and $a_1 + a_2$ are positive integers less than $p$.

From Absolute Value of Integer is not less than Divisors, this is impossible.

Hence by Proof by Contradiction, it is not the case that:

$\exists a_1, a_2: 0 \le a_2 < a_1 \le \dfrac {p - 1} 2: {a_1}^2 = q_1 p + r, {a_2}^2 = q_2 p + r$


To each $r_i$, add $1$ and subtract the result from $p$:

$\forall r_i: s_i = p - \paren {r_i + 1}$

Thus we have $\dfrac {p + 1} 2$ distinct positive integers $s_i$ such that:

$0 \le s_i \le p - 1$

Out of these $r_i$ and $s_i$, there must exist some $r$ and $s$ such that $r = s$, otherwise there would be:

$\dfrac {p + 1} 2 + \dfrac {p + 1} 2 = p + 1$

distinct positive integers less than $p$.

So take such an $r$ and $s$ such that $r = s$.

By construction:

$\exists a, b \in \Z: 0 \le a, b \le \dfrac {p - 1} 2$

such that:

$a^2 = q_1 p + r$
$b^2 = q_2 p + r'$
$s = p - \paren {r' + 1}$

Adding these up:

$a^2 + b^2 + s = \paren {q_1 + q_2 + 1} p + r - 1$

As $r = s$, we can write this as:

$a^2 + b^2 + 1 = m p$

where $m = q_1 + q_2 + 1$.

Thus we have that:

$m p = a^2 + b^2 + 1^2 + 0^2$

and so is a sum of four squares such that:

$m = \dfrac {a^2 + b^2 + 1} p < \dfrac 1 p \paren {\dfrac {p^2} 4 + \dfrac {p^2} 4 + 1} < p$

By hypothesis, $1 \le m < p$.


To recapitulate: what has been proved is that there exists an integer $m$ such that $1 \le m < p$ such that $m p$ is the sum of four squares.

Note the restriction on $m$: of $m = 0$ or $m = p$, then $m p = 0 = 0^2$ or $m p = p^2$, both of which are trivially the sum of four squares.


Every prime $p > 2$ written as sum of $4$ squares

Let $m$ be the smallest positive integer such that:

$(1): \quad m p = {x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2$

for integers $x_1, x_2, x_3, x_4$.

It has already been demonstrated that $m < p$.

It remains to be shown that $m = 1$.


Aiming for a contradiction, suppose $m$ is even.

Let $(1)$ be written as:

$(2): \quad \dfrac m 2 p = \paren {\dfrac {x_1 + x_2} 2}^2 + \paren {\dfrac {x_1 - x_2} 2}^2 + \paren {\dfrac {x_3 + x_4} 2}^2 + \paren {\dfrac {x_3 - x_4} 2}^2$

Then $m p$ is also even.

We have that Parity of Integer equals Parity of its Square.

Thus there are three possibilities:

$\text{(i)}: \quad$ All the $x_i$'s are even.
$\text{(ii)}: \quad$ All the $x_i$'s are odd.
$\text{(iii)}: \quad$ Two of the $x_i$'s are odd, and two of the $x_i$'s are even.

In case $\text{(i)}$ and $\text{(ii)}$ the numbers in parentheses in $(2)$ are integers.

In case $\text{(iii)}$, either $x_1$ and $x_2$ have the same parity or they do not.

If they do, then the numbers in parentheses in $(2)$ are integers.

If they do not, then each of the numbers in parentheses in $(2)$ are odd integers divided by $4$.

Thus $\dfrac m 2 p$ would not be even.

So, given that $\dfrac m 2 p$ is even, it follows that the numbers in parentheses in $(2)$ are integers.

But $\dfrac m 2$ is an integer and $\dfrac m 2 < m$.

This contradicts the statement that $m$ is the smallest positive integer such that $m p$ is the sum of $4$ squares.

By Proof by Contradiction it follows that $m$ is odd.


Aiming for a contradiction, suppose $m \ge 3$.

Again, let $(1)$ be written as:

$(2): \quad \dfrac m 2 p = \paren {\dfrac {x_1 + x_2} 2}^2 + \paren {\dfrac {x_1 - x_2} 2}^2 + \paren {\dfrac {x_3 + x_4} 2}^2 + \paren {\dfrac {x_3 - x_4} 2}^2$

Divide each $x_i$ by $m$ to obtain a remainder $r_i$ such that $0 \le r_i \le m - 1$.

Define $y_i$ as:

$y_i := \begin{cases}

r_i & : 0 \le r_i \le \dfrac {m - 1} 2 \\ r_i - m & : \dfrac {m + 1} 2 \le r_i \le m - 1 \end{cases}$

Then:

$x_i = q_i m + y_i$

where:

$-\dfrac {m - 1} 2 \le y_i \le \dfrac {m - 1} 2$

We have that $y_i = x_i - q_i m$.


Hence $(1)$ gives:

\(\ds \) \(\) \(\ds {y_1}^2 + {y_2}^2 + {y_3}^2 + {y_4}^2\)
\(\ds \) \(=\) \(\ds {x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds 2 m \paren {x_1 q_1 + x_2 q_2 + x_3 q_3 + x_4 q_4}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds m^2 \paren { {q_1}^2 + {q_2}^2 + {q_3}^2 + {q_4}^2}\)
\(\ds \) \(=\) \(\ds m p - 2 m \paren {x_1 q_1 + x_2 q_2 + x_3 q_3 + x_4 q_4}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds m^2 \paren { {q_1}^2 + {q_2}^2 + {q_3}^2 + {q_4}^2}\)
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds m n\)

where $n \in \Z_{\ge 0}$.


Aiming for a contradiction, suppose $n = 0$.

Then all the $y$'s would be zero.

Then all the $x$'s would be divisible by $m$ and so:

$m \paren {\paren {\dfrac {x_1} m}^2 + \paren {\dfrac {x_2} m}^2 + \paren {\dfrac {x_3} m}^2 + \paren {\dfrac {x_4} m}^2} = p$

which means $m \divides p$.

But this is impossible, as $1 < m < p$ and $p$ is prime.

Thus by Proof by Contradiction, $n \ne 0$.


We also have:

\(\ds m n\) \(=\) \(\ds {y_1}^2 + {y_2}^2 + {y_3}^2 + {y_4}^2\)
\(\ds \) \(<\) \(\ds 4 \paren {\dfrac m 2}^2\)
\(\ds \) \(=\) \(\ds m^2\)

and so $n < m$.


Multiplying $(1)$ by $(3)$:

\(\text {(4)}: \quad\) \(\ds m^2 n p\) \(=\) \(\ds \paren { {x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2} \paren { {y_1}^2 + {y_2}^2 + {y_3}^2 + {y_4}^2}\)
\(\ds \) \(=\) \(\ds \paren {x_1 y_1 + x_2 y_2 + x_3 y_3 + x_4 y_4}^2\) Product of Sums of Four Squares
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {x_1 y_2 - x_2 y_1 - x_3 y_4 + x_4 y_3}^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {x_1 y_3 + x_2 y_4 - x_3 y_1 - x_4 y_2}^2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {x_1 y_4 - x_2 y_3 + x_3 y_2 - x_4 y_1}^2\)

Each of the squared numbers on the right hand side is a multiple of $m$, as can be shown for example:

\(\ds \) \(\) \(\ds x_1 y_1 + x_2 y_2 + x_3 y_3 + x_4 y_4\)
\(\ds \) \(=\) \(\ds x_1 \paren {x_1 - q_1 m} + x_2 \paren {x_2 - q_2 m} + x_3 \paren {x_3 - q_3 m} + x_4 \paren {x_4 - q_4 m}\)
\(\ds \) \(=\) \(\ds \paren { {x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2} - m \paren {x_1 q_1 + x_2 q_2 + x_3 q_3 + x_4 q_4}\)
\(\ds \) \(=\) \(\ds m p - m \paren {x_1 q_1 + x_2 q_2 + x_3 q_3 + x_4 q_4}\)
\(\ds \) \(=\) \(\ds m z_1\)


and:

\(\ds \) \(\) \(\ds x_1 y_2 - x_2 y_1 - x_3 y_4 + x_4 y_3\)
\(\ds \) \(=\) \(\ds x_1 \paren {x_2 - q_2 m} - x_2 \paren {x_1 - q_1 m} - x_3 \paren {x_4 - q_4 m} + x_4 \paren {x_3 - q_3 m}\)
\(\ds \) \(=\) \(\ds m \paren {-x_1 q_1 + x_2 q_1 + x_3 q_4 + x_4 q_3}\)
\(\ds \) \(=\) \(\ds m p - m \paren {x_1 q_1 + x_2 q_2 + x_3 q_3 + x_4 q_4}\)
\(\ds \) \(=\) \(\ds m z_2\)

where $z_1$ and $z_2$ are integers

Similarly:

\(\ds x_1 y_3 + x_2 y_4 - x_3 y_1 - x_4 y_2\) \(=\) \(\ds m z_3\)
\(\ds x_1 y_4 - x_2 y_3 + x_3 y_2 - x_4 y_1\) \(=\) \(\ds m z_4\)

for some integers $z_3$ and $z_4$.

Substituting these $m z_1$, $m z_2$, $m z_3$, $m z_4$ back into $(4)$ and dividing by $m^2$ gives:

$n p = \dfrac {\paren {m z_1}^2} {m^2} + \dfrac {\paren {m z_2}^2} {m^2} + \dfrac {\paren {m z_3}^2} {m^2} + \dfrac {\paren {m z_4}^2} {m^2} = {z_1}^2 + {z_2}^2 + {z_3}^2 + {z_4}^2$

where $1 \le n < m$.

But this contradicts the minimality of $m$.

Thus by Proof by Contradiction, $m < 3$.


It remains that $m = 1$ and so $p$ can be expressed as the sum of $4$ squares.

$\blacksquare$


Sources

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