Because $a$ is odd, we can write: :$a = 2 k + 1$ for some positive integer $k$. Then: {{begin-eqn}} {{eqn | l = 4 a - b^2       | r = 4 \paren {2 k + 1} - b^2 }} {{eqn | o = \equiv       | r = 8 k + 4 - 1       | rr = \pmod 8       | c = Odd Square Modulo 8 }} {{eqn | o = \equiv       | r = 3       | rr = \pmod 8 }} {{end-eqn}} Because $b^2 < 4 a$, we have that $4 a - b^2$ is a positive integer.  By Integer as Sum of Three Odd Squares, there exist $3$ odd integers $x, y, z$ such that: :$4 a - b^2 = x^2 + y^2 + z^2$ Because $b, x, y, z$ are all odd integers, $b + x + y + z$ must be even. It is now to be shown that $b + x + y \pm z$ is divisible by $4$. Suppose that $b + x + y + z$ is not divisible by $4$. Because $b + x + y + z$ is even: :$b + x + y + z \equiv 2 \pmod 4$ Writing $z = 2 l + 1$: {{begin-eqn}} {{eqn | l = b + x + y - z       | r = b + x + y + z - 2 z }} {{eqn | o = \equiv       | r = 2 - 2 \paren {2 l + 1}       | rr = \pmod 4 }} {{eqn | o = \equiv       | r = 2 - 4 l - 2        | rr = \pmod 4 }} {{eqn | o = \equiv       | r = 0       | rr = \pmod 4 }} {{end-eqn}} That is: :if $b + x + y + z$ is not divisible by $4$ :then $b + x + y - z$ ''is'' divisible by $4$. {{qed|lemma}} Let us choose the case such that $b + x + y \pm z$ is divisible by $4$. We define: {{begin-eqn}} {{eqn | l = s       | m = \frac {b + x + y \pm z} 4 }} {{eqn | l = t       | m = \frac {b + x - y \mp z} 4       | ro = =       | r = \frac {b + x} 2 - s }} {{eqn | l = u       | m = \frac {b - x + y \mp z} 4       | ro = =       | r = \frac {b + y} 2 - s }} {{eqn | l = v       | m = \frac {b - x - y \pm z} 4       | ro = =       | r = \frac {b \pm z} 2 - s }} {{end-eqn}} We are to show that $s, t, u, v$ are non-negative, and will satisfy: {{begin-eqn}} {{eqn | n = 1       | l = a       | r = s^2 + t^2 + u^2 + v^2 }} {{eqn | n = 2       | l = b       | r = s + t + u + v }} {{end-eqn}} First we show that $s, t, u, v$ are non-negative. Because $x, y, z$ are positive: :$s, t, u, v \ge \dfrac {b - x - y - z} 4$ So we need to show: :$\dfrac {b - x - y - z} 4 \ge 0$ or equivalently: :$\dfrac {b - x - y - z} 4 > -1$ Now: {{begin-eqn}} {{eqn | l = x + y + z       | r = \sqrt {\paren {x + y + z}^2} }} {{eqn | r = \sqrt {x^2 + y^2 + z^2 + 2 x y + 2 y z + 2 z x} }} {{eqn | o = \le       | r = \sqrt {x^2 + y^2 + z^2 + 2 \paren {\sqrt {x^2 + y^2 + z^2} } \paren {\sqrt {y^2 + z^2 + x^2} } }       | c = Cauchy's Inequality }} {{eqn | r = \sqrt {3 \paren {x^2 + y^2 + z^2} } }} {{eqn | r = \sqrt {3 \paren {4 a - b^2} }       | c = $4 a - b^2 = x^2 + y^2 + z^2$ }} {{eqn | o = <       | r = \sqrt {4 \paren {b^2 + 2 b + 4} - 3 b^2}       | c = $3 a < b^2 + 2 b + 4$ }} {{eqn | r = \sqrt {b^2 + 8 b + 16} }} {{eqn | r = b + 4       | c = Square of Sum }} {{eqn | ll = \leadsto       | l = \frac {b - x - y - z} 4       | o = >       | r = -1 }} {{end-eqn}} showing that $s, t, u, v$ are non-negative. Now we check $(1)$: {{begin-eqn}} {{eqn | l = s^2 + t^2 + u^2 + v^2       | r = \paren {\frac {b + x + y \pm z} 4}^2 + \paren {\frac {b + x - y \mp z} 4}^2 + \paren {\frac {b - x + y \mp z} 4}^2 + \paren {\frac {b - x - y \pm z} 4}^2 }} {{eqn | r = \frac 1 {16} ( b^2 + x^2 + y^2 + z^2 + 2 b x + 2 b y \pm 2 b z + 2 x y \pm 2 x z \pm 2 y z }} {{eqn | o =       | r = + \quad b^2 + x^2 + y^2 + z^2 + 2 b x - 2 b y \mp 2 b z - 2 x y \mp 2 x z \pm 2 y z }} {{eqn | o =       | r = + \quad b^2 + x^2 + y^2 + z^2 - 2 b x + 2 b y \mp 2 b z - 2 x y \pm 2 x z \mp 2 y z }} {{eqn | o =       | r = + \quad b^2 + x^2 + y^2 + z^2 - 2 b x - 2 b y \pm 2 b z + 2 x y \mp 2 x z \mp 2 y z ) }} {{eqn | r = \frac 1 {16} \paren {4 \paren {b^2 + x^2 + y^2 + z^2} } }} {{eqn | r = \frac {4 a} 4 }} {{eqn | r = a }} {{end-eqn}} Now we check $(2)$: {{begin-eqn}} {{eqn | l = s + t + u + v       | r = \frac {b + x + y \pm z} 4 + \frac {b + x - y \mp z} 4 + \frac {b - x + y \mp z} 4 + \frac {b - x - y \pm z} 4 }} {{eqn | r = \frac {4 b} 4 + \frac {2 x - 2 x} 4 + \frac {2 y - 2 y} 4 \pm \frac {2 z - 2 z} 4 }} {{eqn | r = b }} {{end-eqn}} Therefore $s, t, u, v$ as defined are the integers we are looking for. {{qed}} {{Namedfor|Augustin Louis Cauchy|cat = Cauchy}}