nicetry007 wrote: Let $ n = s^{2}+t^{2}= u^{2}+v^{2}$. Since $ s \;>\; u$ and $ v\;\geq \; 0$, we have $ v \;>\; 0$, which , in turn, implies that $ u \;>\; 0$ as $ u\;\geq \;v\;$. $ n^{2}\;=\; (s^{2}+t^{2})\; ( u^{2}+v^{2}) \;=\; (su-tv)^{2}\;+\;(sv+tu)^{2}\; = \; (su+tv)^{2}\;+\;(sv-tu)^{2}$ $ \Rightarrow \; |su-tv| \leq n\;,\; |su+tv| \leq n$ (i) $ s^{2}u^{2}\;-\; t^{2}v^{2}\;= \;s^{2}u^{2}\;+\; t^{2}u^{2}\;-\; t^{2}u^{2}\;-\; t^{2}v^{2}\; =\; (s^{2}\;+\; t^{2})\;u^{2}\;-\; (u^{2}\;+\;v^{2})\;t^{2}\;= \;nu^{2}\;-\; nt^{2}\; =\; 0 (mod \; n)$ $ \Rightarrow n \;|\; (su \;+\; tv)(su\;-\; tv)$ (ii) If we strengthen (i) by showing that $ 1 \;\leq \; |su-tv| \;< \;n$ and $ 1 \;\leq \; |su+tv| \;< \;n$, then it follows that $ n$ cannot divide $ |su-tv|$ or $ |su+tv|$. This along with (ii) would imply that $ 1 \;< \;gcd(su-tv \;, \;n)\;<\; n$. In other words, $ gcd(su-tv \;, \;n)$ is a proper divisor of $ n$. Case(a) Suppose $ su \;-\; tv \;=\; 0$. Since $ s \;\geq \; t\; \geq \;0$ and $ u \;\geq \; v\;>\;0$, we have $ su \; \geq \; tv\;\geq \;0 \;\Rightarrow \; su \;-\; tv\; \geq \; 0$. But $ su \;-\; tv = 0$. Hence, we have $ s\; =\; t$ and $ u\; = \; v \;\Rightarrow \; n\; = \;2s^{2}\; =\; 2u^{2}\; \Rightarrow \;s\; = \;u\;$, which is a contradiction. Case(b) Suppose $ su \;+\; tv \;=\; 0$. Since $ s\; \geq \; t\; \geq \;0$ and $ s \;>\; u\; \geq \;v\;> \;0\;$, we have $ su \;+\; tv \;>\; 0$, which contradicts $ su \;+\; tv \;=\; 0$. Case(c) Suppose $ |\;su \;-\; tv \;|\;=\; n$. This implies $ sv+tu \;= \;0$. Since $ s \;\geq \; t\; \geq \;0$ and $ s \;>\;u \;\geq \; v\;> \;0$, we have $ sv+tu \;> \;0$ which contradicts $ sv+tu \;= \;0$. Case(d) Suppose $ |\;su \;+\; tv \;|\;=\; n$. This implies $ sv-tu \;= \;0$. Since $ s \;>\; u\; >\;0$ and $ v \;>\;0$, it follows that $ t \;>\; v\; >\;0 \;\Rightarrow \; s^{2}+t^{2}\;> \; u^{2}+v^{2}$, which contradicts $ s^{2}+t^{2}\;= \; u^{2}+v^{2}$. Hence, we have $ 1 \;\leq \; |su-tv| \;< \;n$ and $ 1 \;\leq \; |su+tv| \;< \;n$. Therefore, it follows that the $ gcd(su-tv \;, \;n)$ is a proper divisor of $ n$.

Alternative Solution: Using Gaussian Integers Let $\alpha:=s+it$ and $\beta:=u+iv$. Thus, $\alpha\beta=(su-tv)+i(sv+tu)$. Because $\alpha\bar{\alpha}=n=\beta\bar{\beta}$, we have \[n^2=(\alpha\beta)(\overline{\alpha\beta}) \geq (su-tv)^2\,.\] If $|su-tv|$ were equal to $n$, then $sv+tu=0$, which cannot be true as $s>u\geq v >0$. Therefore, $|su-tv|<n$, and we arrive at the conclusion.

If the assumptions on $s, t, u, v$ were instead $|s| \geq |t| \geq 0$, $|u| \geq |v| \geq 0$ and $|s| > |u|$, then Batominovski's solution would still work with minor modification. In the original form, one could also argue as follows: $s>u$ implies that $u \geq v > t$. Therefore, $su-tv > 0$. If $n$ divides $su-tv$, then $n \leq su-tv \leq su < s^{2} \leq n$, which is a contradiction.