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Question Number 157163 by MathSh last updated on 20/Oct/21

$$\mathrm{Find}\mathrm{the}\mathrm{positive}\mathrm{integer}\mathrm{solution}$$$$\mathrm{of}\mathrm{the}\mathrm{equation}:$$$${\mathrm{x}}^3+{\mathrm{y}}^3=7\cdot (14\mathrm{xy}+3)$$

Commented by Rasheed.Sindhi last updated on 22/Oct/21

$${\mathrm{x}}^3+{\mathrm{y}}^3=7\cdot (14\mathrm{xy}+3)$$$${\mathrm{x}}^3+{\mathrm{y}}^3-21=7^2.2\mathrm{xy}$$$${\mathrm{x}}^3+{\mathrm{y}}^3-21\in \mathbb{E}$$$$\Rightarrow {\mathrm{x}}^3+{\mathrm{y}}^3\in \mathbb{O}$$$$\Rightarrow \mathrm{one}\mathrm{of}\mathrm{x},\mathrm{y}\mathrm{is}\mathrm{odd}\mathrm{and}\mathrm{other}\mathrm{is}\mathrm{even}.$$$$\mathrm{As}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{are}\mathrm{exchangeable}\mathrm{we}\mathrm{may}$$$$\mathrm{take}\mathrm{any}\mathrm{one}\mathrm{of}\mathrm{these}\mathrm{as}\mathrm{odd}.$$$$\mathrm{let}\mathrm{x}\in \mathbb{O},\mathrm{y}\in \mathbb{E}$$$${(2\mathrm{m}+1)}^3+{\left(2\mathrm{n}\right)}^3-21=98(2\mathrm{m}+1)\left(2\mathrm{n}\right)$$$$▸{8\mathrm{m}}^3+1+6\mathrm{m}(2\mathrm{m}+1)+{8\mathrm{n}}^3-21$$$$=98(2\mathrm{m}+1)\left(2\mathrm{n}\right)$$$$▸{8\mathrm{m}}^3+{8\mathrm{n}}^3+6\mathrm{m}(2\mathrm{m}+1)-20$$$$=98(2\mathrm{m}+1)\left(2\mathrm{n}\right)$$$$▸{4\mathrm{m}}^3+{4\mathrm{n}}^3+3\mathrm{m}(2\mathrm{m}+1)-10$$$$=49(2\mathrm{m}+1)\left(2\mathrm{n}\right)$$$$49(2\mathrm{m}+1)\left(2\mathrm{n}\right)\in \mathbb{E}$$$$\Rightarrow {4\mathrm{m}}^3+{4\mathrm{n}}^3+3\mathrm{m}(2\mathrm{m}+1)-10\in \mathbb{E}$$$$\Rightarrow 3\mathrm{m}(2\mathrm{m}+1)\in \mathbb{E}$$$$\Rightarrow {6\mathrm{m}}^2+3\mathrm{m}\in \mathbb{E}\Rightarrow \mathrm{m}\in \mathbb{E}$$$$\mathrm{x}=2\mathrm{m}+1=2\left(2\mathrm{k}\right)+1=4\mathrm{k}+1;\mathrm{k}\in \mathbb{W}$$$$\mathrm{x}=1,5,9,....$$$$\mathrm{x}=1:$$$${\left(1\right)}^3+{\mathrm{y}}^3=7(14.1.\mathrm{y}+3)$$$${\mathrm{y}}^3-98\mathrm{y}-20=0$$$$(\mathrm{y}-10)\left(\underset{\mathrm{y}\notin \mathbb{N}}{\underbrace{{\mathrm{y}}^2+10\mathrm{y}+2}}\right)=0$$$$\mathrm{y}=10$$$$\mathrm{Continue}$$

Answered by Rasheed.Sindhi last updated on 20/Oct/21

$${\mathrm{x}}^3+{\mathrm{y}}^3=7\cdot (14\mathrm{xy}+3);\stackrel{?}{\mathrm{x}},\stackrel{?}{\mathrm{y}}\in {\mathbb{Z}}^{+}$$$${\left(\underset{\mathrm{u}}{\underbrace{\mathrm{x}+\mathrm{y}}}\right)}^3-3\underset{\mathrm{v}}{\underbrace{\mathrm{xy}}}(\mathrm{x}+\mathrm{y})=7\cdot (14\mathrm{xy}+3)$$$${\mathrm{u}}^3-3\mathrm{uv}=7(14\mathrm{v}+3)$$$$\mathrm{u}({\mathrm{u}}^2-3\mathrm{v})=7(14\mathrm{v}+3)$$$$\mathrm{case}1:\mathrm{u}=7\wedge {\mathrm{u}}^2-3\mathrm{v}=14\mathrm{v}+3$$$$49-3\mathrm{v}-14\mathrm{v}-3=0$$$$17\mathrm{v}=46\Rightarrow \mathrm{v}\notin {\mathbb{Z}}^{+}.........\left(\mathrm{i}\right)$$$$\mathrm{case}2:{\mathrm{u}}^2-3\mathrm{v}=7\wedge \mathrm{u}=14\mathrm{v}+3$$$${(14\mathrm{v}+3)}^2-3\mathrm{v}=7$$$${196\mathrm{v}}^2+84\mathrm{v}+9-3\mathrm{v}-7=0$$$${196\mathrm{v}}^2+81\mathrm{v}+2=0$$$$\mathrm{v}\notin {\mathbb{Z}}^{+}......................\left(\mathrm{ii}\right)$$$$\mathrm{In}\mathrm{both}\mathrm{possible}\mathrm{cases}:$$$$\mathrm{v}\notin {\mathbb{Z}}^{+}\Rightarrow \mathrm{xy}\notin {\mathbb{Z}}^{+}\Rightarrow \mathrm{x}\notin {\mathbb{Z}}^{+}\vee \mathrm{y}\notin {\mathbb{Z}}^{+}$$$$\mathrm{Solution}\mathrm{not}\mathrm{possible}!$$

Commented by MathSh last updated on 21/Oct/21

$$\mathrm{Thank}\mathrm{you}\mathrm{dear}\mathrm{Ser},\mathrm{but}\mathrm{ans}:(1;10)(10;1)$$

Commented by Rasheed.Sindhi last updated on 21/Oct/21

$$\mathbb{T}\mathrm{hank}\mathrm{you}\mathrm{dear}\stackrel{?}{\mathrm{sir}}/\stackrel{?}{\mathrm{madam}}.$$$$\mathrm{Certainly}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{logical}\mathrm{error}\mathrm{and}$$$${\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{think}\mathrm{over}\mathrm{it}.$$$$$$

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