Proof.

Without loss of generality, let \(2 \nmid a\). By construction, the following congruences hold. \begin{align*} a^2 \equiv 1 \pmod {4}&, \quad b^2 \equiv 0 \pmod {4} \\ c^2 \equiv a^2 + b^2 &\equiv 1 \pmod {4} \end{align*}

Therefore, \(c\) is odd. Let \(a = 2a' + 1\) and \(c = 2c' + 1\). Then, the following congruences hold. \begin{align*} (2a' + 1)^2 + b^2 &\equiv (2c' + 1) \\ 4a'^2 + 4a' + 1 + b^2 &\equiv 4c'^2 + 4c' + 1 \pmod {8} \\ b^2 \equiv 4c'^2 + 4c' - 4a'^2 - 4a' &\equiv 4 (c'^2 + c' - a'^2 - a') \equiv 0 \pmod {8} \end{align*}

Thus \(4 \mid b\). Continuing, the following equation holds by definition. \[ (c + a)(c - a) = b^2 \] Let \(d = (c + a, c - a)\). Note that \(2 \mid d\) since \(2 \nmid a, c\). Moreover, \(d \mid (c + a) + (c - a) = 2c\) and \(d \mid (c + a) - (c - a) = 2a\). Because \((a, c) = 1\), \(d = 2\). In other words, \(c + a = 2m^2\) and \(c - a = 2n^2\) for some coprime positive integers \(m, n\). Substituting, \(b = 2mn\). Thus, there exists a pair of positive integers \(m > n\) such that \(\{ a, b \} = \{ m^2 - n^2, 2mn \}\) and \(c = m^2 + n^2\).

When using this theorem, it is very important to consider the condition stated that \((a, b) = (b, c) = (c, a) = 1\). Moreover, it is important to consider the parity of \(a\) and \(b\) before assigning respective values \(m^2 - n^2\) and \(2mn\). Lastly, but most importantly, remember that \((m, n) = 1\) and they are opposite parity! The last condition is very easy to miss (at least for me) and used very often when we apply this theorem recursively.

Solution.

First, let \((x, y)\) be another coprime pair such that \(x^2 - 5y^2 = m^2\) and \(x^2 + 5y^2 = n^2\) for some integers \(m\) and \(n\). Rearranging the equation, \(2x^2 = m^2 + n^2\) and \(10y^2 = n^2 - m^2\) hold. Starting with the first equation, the following equation is obtained. \[ x^2 = \frac {m^2 + n^2}{2} = \left ( \frac {m + n}{2} \right )^2 + \left ( \frac {m - n}{2} \right )^2 \] From the equation, assume that \(x\), \(\frac {m + n}{2}\), and \(\frac {m - n}{2}\) are pairwise coprime. Note that we are allowed to set this condition since we just need to find any \(x\) and \(y\). By Theorem 1.1.1 , there exists coprime integers \(u, v\) with opposite parity such that \(x = u^2 + v^2\), \(\frac {m - n}{2} = u^2 - v^2\), and \(\frac {m + n}{2} = 2uv\). Again, we are allowed to set parities since we are finding any \(x\) and \(y\).

Continuing, the second equation can be modified. \begin{align*} 10y^2 &= 4 \cdot \frac {n + m}{2} \cdot \frac {n - m}{2} \\ \frac {5}{2} y^2 &= (u^2 - v^2) 2uv \\ \therefore 5 \left ( \frac {y}{2} \right )^2 &= (u + v)(u - v) uv \end{align*}

By construction, \(u, v\) are coprime to each other and \((u + v)(u - v)\) and \(uv\) are also coprime. Let \(u = 41^2\) and \(v = 5 \cdot 12^2\). Notice that the conditions that we have set while defining the variables are met. Moreover as given by the problem, it is evident that \((41^2 + 5 \cdot 12^2)(41^2 - 5 \cdot 12^2)\) are perfect squares, returning an integer \(y\). Hence, the following equations are obtained. \begin{align*} 5 \left ( \frac {y}{2} \right )^2 &= (41^2 + 5 \cdot 12^2)(41^2 - 5 \cdot 12^2) \cdot 41^2 \cdot 5 \cdot 12^2 \\ \left ( \frac {y}{2} \right )^2 &= (41^2 + 5 \cdot 12^2)(41^2 - 5 \cdot 12^2) \cdot 41^2 \cdot 12^2 \\ \therefore y &= 2 \cdot 12 \cdot 41 \cdot \sqrt {(41^2 + 5 \cdot 12^2)(41^2 - 5 \cdot 12^2)} = 1494696 \end{align*}

Moreover, \(x = u^2 + v^2 = \left (41^2\right )^2 + \left (5 \cdot 12^2\right )^2 = 3344161\). Therefore, \((3344161, 1494696)\) is another pair.