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Question Number 150558 by mathdanisur last updated on 13/Aug/21

Solve for integers:  (6x + 5y^2 )(4z + x)(2y^2  + 3z) = 2021

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{integers}: \\ $$$$\left(\mathrm{6x}\:+\:\mathrm{5y}^{\mathrm{2}} \right)\left(\mathrm{4z}\:+\:\mathrm{x}\right)\left(\mathrm{2y}^{\mathrm{2}} \:+\:\mathrm{3z}\right)\:=\:\mathrm{2021} \\ $$

Answered by MJS_new last updated on 13/Aug/21

2021=43×47  abc=2021  one factor =1  one factor =43  one factor =47    (1) 6x+5y^2 =a ⇒ y^2 =((a−6x)/5)  (2) 4z+x=b  (3) 2y^2 +3z=c ⇒ y^2 =((c−3z)/2)  ================  (1)/(3)  12x−15z−2a+5c=0           (2)        x+   4z−b            =0  ⇒  x=((8a+15b−20c)/(63))∧z=((−2a+12b+5c)/(63))  ⇒  y=±(√((a−6b+8c)/(21)))  testing all variations for a, b, c = 1, 43, 47  ⇒  no integer solution

$$\mathrm{2021}=\mathrm{43}×\mathrm{47} \\ $$$${abc}=\mathrm{2021} \\ $$$$\mathrm{one}\:\mathrm{factor}\:=\mathrm{1} \\ $$$$\mathrm{one}\:\mathrm{factor}\:=\mathrm{43} \\ $$$$\mathrm{one}\:\mathrm{factor}\:=\mathrm{47} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{6}{x}+\mathrm{5}{y}^{\mathrm{2}} ={a}\:\Rightarrow\:{y}^{\mathrm{2}} =\frac{{a}−\mathrm{6}{x}}{\mathrm{5}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{4}{z}+{x}={b} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2}{y}^{\mathrm{2}} +\mathrm{3}{z}={c}\:\Rightarrow\:{y}^{\mathrm{2}} =\frac{{c}−\mathrm{3}{z}}{\mathrm{2}} \\ $$$$================ \\ $$$$\left(\mathrm{1}\right)/\left(\mathrm{3}\right)\:\:\mathrm{12}{x}−\mathrm{15}{z}−\mathrm{2}{a}+\mathrm{5}{c}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:{x}+\:\:\:\mathrm{4}{z}−{b}\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}=\frac{\mathrm{8}{a}+\mathrm{15}{b}−\mathrm{20}{c}}{\mathrm{63}}\wedge{z}=\frac{−\mathrm{2}{a}+\mathrm{12}{b}+\mathrm{5}{c}}{\mathrm{63}} \\ $$$$\Rightarrow \\ $$$${y}=\pm\sqrt{\frac{{a}−\mathrm{6}{b}+\mathrm{8}{c}}{\mathrm{21}}} \\ $$$$\mathrm{testing}\:\mathrm{all}\:\mathrm{variations}\:\mathrm{for}\:{a},\:{b},\:{c}\:=\:\mathrm{1},\:\mathrm{43},\:\mathrm{47} \\ $$$$\Rightarrow \\ $$$$\mathrm{no}\:\mathrm{integer}\:\mathrm{solution} \\ $$

Commented by mathdanisur last updated on 14/Aug/21

Thank you Ser cool

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser}\:\mathrm{cool} \\ $$

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