sokve x^3 +6y^3 =4z^3 x y z integers


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Question Number 35297 by 123 45 polytechnicien last updated on 17/May/18

$s o k v e x^{3} + 6 y^{3} = 4 z^{3} x y z i n t e g e r s$
Commented by Rasheed.Sindhi last updated on 18/May/18

${4 z}^{3} \to even \Rightarrow x^{3} + {6 y}^{3} \to even$ $x^{3} + {6 y}^{3} \to even \land {6 y}^{3} \to even$ $\Rightarrow x^{3} \to even \Rightarrow x \to even$ $Let x = 2 m$ $x^{3} + 6 y^{3} = 4 z^{3} \Rightarrow {8 m}^{3} + {6 y}^{3} = {4 z}^{3}$ $\Rightarrow {4 m}^{3} + {3 y}^{3} = {2 z}^{3}$ ${4 m}^{3} \to even \land {2 z}^{3} \to even \Rightarrow {3 y}^{3} \to even$ $\Rightarrow y^{3} \to even \Rightarrow y \to even$ $x^{3} + 6 y^{3} = {4 z}^{3} \Rightarrow {(even)}^{3} + 6 {(even)}^{3} = {4 z}^{3}$ $8 ∣ {(even)}^{3} \Rightarrow 8 ∣ [{(even)}^{3} + 6 {(even)}^{3}]$ $\Rightarrow 8 ∣ {4 z}^{3} \Rightarrow z^{3} \to even \Rightarrow z \to even$ $So x, y, z \to even$ $Let x = 2 m, y = 2 n & z = 2 l$ $x^{3} + 6 y^{3} = 4 z^{3} \Rightarrow {(2 m)}^{3} + 6 {(2 n)}^{3} = 4 {(2 l)}^{3}$ ${8 m}^{3} + 6 ({8 n}^{3}) = 4 (8 l^{3})$ $m^{3} + {6 n}^{3} = 4 l^{3}$ $This is same as x^{3} + 6 y^{3} = 4 z^{3}$ $(hahaha we reeched at the begning)$ $Anyway we should search x, y, z in$ $in even numbers and this is a clue!$
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