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Question Number 19329 by Tinkutara last updated on 09/Aug/17

Solve the equation y^3  = x^3  + 8x^2  − 6x + 8  for positive integers x and y.

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:{y}^{\mathrm{3}} \:=\:{x}^{\mathrm{3}} \:+\:\mathrm{8}{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:\mathrm{8} \\ $$$$\mathrm{for}\:\mathrm{positive}\:\mathrm{integers}\:{x}\:\mathrm{and}\:{y}. \\ $$

Commented by Tinkutara last updated on 12/Aug/17

RMO 2000

$$\mathrm{RMO}\:\mathrm{2000} \\ $$

Commented by Tinkutara last updated on 12/Aug/17

help pls

$$\mathrm{help}\:\mathrm{pls} \\ $$

Commented by Tinkutara last updated on 12/Aug/17

How is (x + 3)^3  − y^3  = x^3  + 9x^2  + 27x  + 27 − (x^3  + 8x^2  − 6x + 8) = x^2  + 33x  + 19 > 0?

$$\mathrm{How}\:\mathrm{is}\:\left({x}\:+\:\mathrm{3}\right)^{\mathrm{3}} \:−\:{y}^{\mathrm{3}} \:=\:{x}^{\mathrm{3}} \:+\:\mathrm{9}{x}^{\mathrm{2}} \:+\:\mathrm{27}{x} \\ $$$$+\:\mathrm{27}\:−\:\left({x}^{\mathrm{3}} \:+\:\mathrm{8}{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:\mathrm{8}\right)\:=\:{x}^{\mathrm{2}} \:+\:\mathrm{33}{x} \\ $$$$+\:\mathrm{19}\:>\:\mathrm{0}? \\ $$

Commented by ajfour last updated on 12/Aug/17

given:  y=x^3 +8x^2 −6x+8       (x+3)^3 =x^3 +9x^2 +27x+27  So    (x+3)^3 −y^3 =x^2 +33x+19  for positive integer x, R.H.S. ≥ 53

$$\mathrm{given}:\:\:\mathrm{y}=\mathrm{x}^{\mathrm{3}} +\mathrm{8x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{8} \\ $$$$\:\:\:\:\:\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{3}} =\mathrm{x}^{\mathrm{3}} +\mathrm{9x}^{\mathrm{2}} +\mathrm{27x}+\mathrm{27} \\ $$$$\mathrm{So}\:\:\:\:\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} =\mathrm{x}^{\mathrm{2}} +\mathrm{33x}+\mathrm{19} \\ $$$$\mathrm{for}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{x},\:\mathrm{R}.\mathrm{H}.\mathrm{S}.\:\geqslant\:\mathrm{53} \\ $$

Commented by Tinkutara last updated on 12/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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