$1$ 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: :$\forall r_i: 0 \le r_i \le p - 1$ {{AimForCont}}: :$\exists a_1, a_2: 0 \le a_2 < a_1 \le \dfrac {p - 1} 2: {a_1}^2 = q_1 p + r, {a_2}^2 = q_2 p + r$ That is, two different integers in that range which have the same remainder. Then: {{begin-eqn}} {{eqn | l = {a_1}^2 - {a_2}^2       | r = \paren {q_1 - q_2} p       | c =  }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = \paren { {a_1}^2 - {a_2}^2}       | c =  }} {{eqn | r = \paren {a_1 - a_2} \paren {a_1 + a_2}       | c =  }} {{end-eqn}} 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 $m = 1$. {{AimForCont}} $m$ 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. {{AimForCont}} $m \ge 3$. 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: {{begin-eqn}} {{eqn | o =        | r = {y_1}^2 + {y_2}^2 + {y_3}^2 + {y_4}^2       | c =  }} {{eqn | r = {x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2       | c =  }} {{eqn | o =        | ro= -       | r = 2 m \paren {x_1 q_1 + x_2 q_2 + x_3 q_3 + x_4 q_4}       | c =  }} {{eqn | o =        | ro= +       | r = m^2 \paren { {q_1}^2 + {q_2}^2 + {q_3}^2 + {q_4}^2}       | c =  }} {{eqn | r = m p - 2 m \paren {x_1 q_1 + x_2 q_2 + x_3 q_3 + x_4 q_4}       | c =  }} {{eqn | o =        | ro= +       | r = m^2 \paren { {q_1}^2 + {q_2}^2 + {q_3}^2 + {q_4}^2}       | c =  }} {{eqn | n = 3       | r = m n       | c =  }} {{end-eqn}} where $n \in \Z_{\ge 0}$. {{AimForCont}} $n = 0$. 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: {{begin-eqn}} {{eqn | l = m n       | r = {y_1}^2 + {y_2}^2 + {y_3}^2 + {y_4}^2       | c =  }} {{eqn | o = <       | r = 4 \paren {\dfrac m 2}^2       | c =  }} {{eqn | r = m^2       | c =  }} {{end-eqn}} and so $n < m$. Multiplying $(1)$ by $(3)$: {{begin-eqn}} {{eqn | n = 4       | l = m^2 n p       | r = \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}       | c =  }} {{eqn | r = \paren {x_1 y_1 + x_2 y_2 + x_3 y_3 + x_4 y_4}^2       | c = Product of Sums of Four Squares }} {{eqn | o =        | ro= +       | r = \paren {x_1 y_2 - x_2 y_1 - x_3 y_4 + x_4 y_3}^2       | c =  }} {{eqn | o =        | ro= +       | r = \paren {x_1 y_3 + x_2 y_4 - x_3 y_1 - x_4 y_2}^2       | c =  }} {{eqn | o =        | ro= +       | r = \paren {x_1 y_4 - x_2 y_3 + x_3 y_2 - x_4 y_1}^2       | c =  }} {{end-eqn}} Each of the squared numbers on the {{RHS}} is a multiple of $m$, as can be shown for example: {{begin-eqn}} {{eqn | o =        | r = x_1 y_1 + x_2 y_2 + x_3 y_3 + x_4 y_4       | c =  }} {{eqn | r = 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}       | c =  }} {{eqn | r = \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}       | c =  }} {{eqn | r = m p - m \paren {x_1 q_1 + x_2 q_2 + x_3 q_3 + x_4 q_4}       | c =  }} {{eqn | r = m z_1       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | o =        | r = x_1 y_2 - x_2 y_1 - x_3 y_4 + x_4 y_3       | c =  }} {{eqn | r = 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}       | c =  }} {{eqn | r = m \paren {-x_1 q_1 + x_2 q_1 + x_3 q_4 + x_4 q_3}       | c =  }} {{eqn | r = m p - m \paren {x_1 q_1 + x_2 q_2 + x_3 q_3 + x_4 q_4}       | c =  }} {{eqn | r = m z_2       | c =  }} {{end-eqn}} where $z_1$ and $z_2$ are integers Similarly: {{begin-eqn}} {{eqn | l = x_1 y_3 + x_2 y_4 - x_3 y_1 - x_4 y_2       | r = m z_3       | c =  }} {{eqn | l = x_1 y_4 - x_2 y_3 + x_3 y_2 - x_4 y_1       | r = m z_4       | c =  }} {{end-eqn}} 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. {{qed}}