Solve for positive integers: a^3 + 9b^2 + 9c^2 = 2017 where a ≥ b ≥ c


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Question Number 159057 by HongKing last updated on 12/Nov/21

$Solve for positive integers :$ $a^{3} + {9 b}^{2} + {9 c}^{2} = 2017$ $where a ⩾ b ⩾ c$
Commented by Rasheed.Sindhi last updated on 12/Nov/21

$(a, b, c) = (10, 8, 7)$
	Answered by Rasheed.Sindhi last updated on 12/Nov/21

	$a^{3} + {9 b}^{2} + {9 c}^{2} = 2017$ $where a ⩾ b ⩾ c \land a, b, c \in Z^{+}$ $b^{2} + c^{2} = \frac{2017 - a^{3}}{9}$ $\Rightarrow 9 ∣ (2017 - a^{3})$ $a ⩽ ⌊ \sqrt[3]{2017} ⌋ = 12$ $a = 12 : 9 ∤ (2017 - 12^{3}) (r e j e c t e d)$ $W e c a n o b s e r v e t h a t o n l y$ $9 ∣ 2017 - 10^{3}$ $9 ∣ 2017 - 7^{3}$ $9 ∣ 2017 - 4^{3}$ $9 ∣ 2017 - 1^{3}$ $S o p o s s i b l e v a l u e s f o r a are :$ $10, 7, 4, 1$ $a = 10$ $b^{2} + c^{2} = \frac{2017 - a^{3}}{9} = \frac{2017 - 10^{3}}{9} = 113$ $N o t e t h a t b & c c a n h a v e v a l u e s u p t o a$ $H e r e b, c ⩽ 10, w e c a n e a s i l y f i n d$ $113 = 8^{2} + 7^{2}$ $a = 7$ $b^{2} + c^{2} = \frac{2017 - 7^{3}}{9} = 186$ $m a x i m u m v a l u e f o r b & c i s 7$ $a n d m a x (b^{2} + c^{2}) = 7^{2} + 7^{2} = 98$ $b^{2} + c^{2} \neq 186$ $S i m i l a r l o g i c s h o w s t h a t f o r a = 4, 1$ $t h e r e a r e n o v a l u e s f o r b, c u n d e r$ $t h e a b o v e c o n d i t i o n s .$ $\underset{⌢}{\overset{\overset{}{⌢}}{\lor \land \lor \land \lor \land \lor \land \lor \land \lor \land \lor \land \lor \land \lor \land \lor \land \lor \land \lor \land}}$ $1 st Filter : \begin{matrix} a ⩽ ⌊ \sqrt[3]{2017} ⌋ = 12 \end{matrix}$ $2 nd Filter : \begin{matrix} 9 ∣ (2017 - a^{3}) \end{matrix}$ $3 rd Filter :$ $\begin{matrix} b, c ⩽ a \land b^{2} + c^{2} = (2017 - a^{3}) / 9 \end{matrix}$
	Commented by HongKing last updated on 14/Nov/21

	$very nice dear Ser thank you$
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