Fie x,y,z,t numere naturale, oricare doua diferite si prime intre ele. Sa se demonstreze ca daca \( \sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{t} \), atunci x,y,z,t sunt patrate perfecte.
C. ,,Gh. Titeica'',2004
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\( \sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{t} \)
\( \Longleftrightarrow \sqrt{xt}+\sqrt{yt}+\sqrt{zt}=t \)
\( \Longleftrightarrow \sqrt{xt}+\sqrt{yt}=t-\sqrt{zt} \)
\( \Longleftrightarrow xt+yt+2t\sqrt{xy}=t^{2}-2t\sqrt{zt}+zt \)
\( \Longleftrightarrow x+y+2\sqrt{xy}=t-2\sqrt{zt}+z \)
\( \Longleftrightarrow \sqrt{xy}=\frac{t+z-x-y-2\sqrt{zt}}{2} \)
\( \Longleftrightarrow \sqrt{xy}=a-\sqrt{zt},\ unde\ a=\frac{t+z-x-y}{2}. \)
Daca \( a=0, \) atunci \( \sqrt{xy}+\sqrt{zt}=0, \) imposibil. Deci \( xy=a^{2}-2a\sqrt{zt}+zt \Longrightarrow \sqrt{zt}=\frac{a^{2}+zt-xy}{2a}\in\mathbb{Q} \) si urmeaza ca \( \sqrt{zt}\in\mathbb{N} \), dar \( (z,\ t)=1 \) si rezulta ca \( z=k^{2},\ t=q^{2},\ k,q\in\mathbb{N}^{\ast}. \) etc.
\( \Longleftrightarrow \sqrt{xt}+\sqrt{yt}+\sqrt{zt}=t \)
\( \Longleftrightarrow \sqrt{xt}+\sqrt{yt}=t-\sqrt{zt} \)
\( \Longleftrightarrow xt+yt+2t\sqrt{xy}=t^{2}-2t\sqrt{zt}+zt \)
\( \Longleftrightarrow x+y+2\sqrt{xy}=t-2\sqrt{zt}+z \)
\( \Longleftrightarrow \sqrt{xy}=\frac{t+z-x-y-2\sqrt{zt}}{2} \)
\( \Longleftrightarrow \sqrt{xy}=a-\sqrt{zt},\ unde\ a=\frac{t+z-x-y}{2}. \)
Daca \( a=0, \) atunci \( \sqrt{xy}+\sqrt{zt}=0, \) imposibil. Deci \( xy=a^{2}-2a\sqrt{zt}+zt \Longrightarrow \sqrt{zt}=\frac{a^{2}+zt-xy}{2a}\in\mathbb{Q} \) si urmeaza ca \( \sqrt{zt}\in\mathbb{N} \), dar \( (z,\ t)=1 \) si rezulta ca \( z=k^{2},\ t=q^{2},\ k,q\in\mathbb{N}^{\ast}. \) etc.