Solve-for-positive-integers-a-3-9b-2-9c-2-2017-where-a-b-c-

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