=== Proof of Existence === There are three possibilities for a prime: :$(1): \quad p = 2$ or: :$(2): \quad p \equiv 1 \pmod 4$ or: :$(3): \quad p \equiv 3 \pmod 4$ ==== Necessary Condition ==== Suppose $p$ can be expressed as the sum of two squares. First we note that $2 = 1^2 + 1^2$, which is the sum of two squares. This disposes of the case where $p = 2$. Let $p = a^2 + b^2$. From Sum of Two Squares not Congruent to 3 modulo 4, $p \not \equiv 3 \pmod 4$ whatever $a$ and $b$ are. So either $p = 2$, or $p \equiv 1 \pmod 4$. ==== Sufficient Condition ==== We have already noted that $2 = 1^2 + 1^2$, which is the sum of two squares. Let $p$ be a prime number of the form $p \equiv 1 \pmod 4$. Suppose $m p = x^2 + y^2$ has a solution such that $1 < m < p$. Let $u, v$ be the least absolute residues modulo $m$ of $x$ and $y$ respectively. That is: :$u \equiv x, v \equiv y \pmod m: \dfrac {-m} 2 < u, v \le \dfrac m 2$ Then: :$u^2 + v^2 \equiv x^2 + y^2 \pmod m$ Thus: :$\exists r \in \Z, r \ge 0: u^2 + v^2 = m r$ We are going to establish a descent step. That is, we aim to show that $r p$ is the sum of two squares with $1 \le r < m$. First we show that $r$ does lie in this range. If $r = 0$ then $u = v = 0$ and so $m$ divides both $x$ and $y$. But then from $m p = x^2 + y^2$ we have that $m \divides p$. This cannot happen as $p$ is prime. So: :$1 \le r = \dfrac {u^2 + v^2} m \le \dfrac 1 m \times \paren {\dfrac {m^2} 4 + \dfrac {n^2} 4} = \dfrac m 2 < m$ So $1 \le r < m$. Now we show that $r p$ is the sum of two squares. Multiplying $m p = x^2 + y^2$ and $m r = u^2 + v^2$: {{begin-eqn}} {{eqn | l = m^2 r p       | r = \paren {x^2 + y^2} \paren {u^2 + v^2}       | c =  }} {{eqn | r = \paren {x u + y v}^2 + \paren {x v - y u}^2       | c = Brahmagupta-Fibonacci Identity }} {{end-eqn}} Now: :$x u + y v \equiv x^2 + y^2 \equiv 0 \pmod m$, so $m \divides x u + y v$ :$x v - y u \equiv x y - x y \equiv 0 \pmod m$, so $m \divides x v - y u$ So, putting $m X = x u + y v, m Y = x v - y u$, we get: :$m^2 r p = m^2 X^2 + m^2 Y^2$ That is: :$r p = X^2 + Y^2$ Hence the descent step is established. Next we need to show that $m p = x^2 + y^2$ has a solution for some $m$ with $1 \le m < p$. From First Supplement to Law of Quadratic Reciprocity, we have that $-1$ is a quadratic residue for each prime $p \equiv 1 \pmod 4$. Hence the congruence $x^2 + 1 \equiv 0 \pmod p$ has a least positive solution $x_1$ such that $1 \le x_1 \le p - 1$. So there exists a positive integer $m$ such that $m p = x_1^2 + 1^2$. This is just what we want, because: :$m = \dfrac {x_1^2 + 1^2} p \le \dfrac {\paren {p - 1}^2 + 1} p = \dfrac {p^2 - 2 \paren {p - 1}^2} p < p$ If this solution has $m > 1$, then our descent step (above) guarantees a solution for a smaller positive value of $m$. Eventually we will reach a solution with $m = 1$, that is: :$p = x^2 + y^2$ {{qed}} === Proof of Uniqueness === Let $p$ be prime such that $p \equiv 1 \pmod 4$. Suppose $p = a^2 + b^2 = c^2 + d^2$ where $a > b > 0$ and $c > d > 0$. We are going to show that $a = c$ and $b = d$. From the two expressions for $p$, we have: {{begin-eqn}} {{eqn | l = \paren {a d - b c} \paren {a d + b c}       | r = a^2 d^2 - b^2 c^2       | c = Difference of Two Squares }} {{eqn | r = \paren {p - b^2} d^2 - b^2 \paren {p - d^2}       | c = substituting for $a^2$ and $c^2$ }} {{eqn | r = p d^2 - b^2 d^2 - p b^2 + b^2 d^2       | c =  }} {{eqn | r = p \paren {d^2 - b^2}       | c =  }} {{eqn | o = \equiv       | r = 0       | c = $\pmod p$ }} {{end-eqn}} So we have: :$\paren {a d - b c} \paren {a d + b c} \equiv 0 \pmod p$ From Euclid's Lemma, that means: :$p \divides \paren {a d - b c}$ or: :$p \divides \paren {a d + b c}$ So, suppose $p \divides \paren {a d + b c}$. Now, we have that each of $a^2, b^2, c^2, d^2$ must be less than $p$. Hence $0 < a, b, c, d < \sqrt p$. This implies $0 < a d + b c < 2p$. That must mean that $a d + b c = p$. But then: {{begin-eqn}} {{eqn | l = p^2       | r = \paren {a^2 + b^2} \paren {d^2 + c^2}       | c =  }} {{eqn | r = \paren {a d + b c}^2 + \paren {a c - b d}^2       | c = Brahmagupta-Fibonacci Identity }} {{eqn | r = p^2 + \paren {a c - b d}^2       | c =  }} {{end-eqn}} That means: :$a c - b d = 0$ But since $a > b$ and $c > d$ we have: :$a c > b d$ This contradiction shows that $a d + b c$ can not be divisible by $p$. So this means: :$p \divides \paren {a d - b c}$ Similarly, because $0 < a, b, c, d < \sqrt p$ we have: :$-p < a d - b c < p$ This means: :$a d = b c$ So: :$a \divides b c$ But $a \perp b$ otherwise $a^2 + b^2$ has a common divisor greater than $1$ and less than $p$. This cannot happen because $p$ is prime. So by Euclid's Lemma: :$a \divides c$ So we can put $c = k a$ and so $a d = b c$ becomes $d = k b$. Hence: :$p = c^2 + d^2 = k^2 \paren {a^2 + b^2} = k^2 p$ This means $k = 1$ which means $a = c$ and $b = d$ as we wanted to show. {{qed}}