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Question Number 184551 by a.lgnaoui last updated on 08/Jan/23

Resoudre dans Z^+ x+y+(√(xy)) =39

ResoudredansZ+x+y+xy=39

Answered by Frix last updated on 08/Jan/23

xy≥0 ⇒ x≥0∧y≥0∨x<0∧y<0 (1) x<0∧y<0 Let p, q >0∧x=−p^2 ∧y=−q^2 ⇒ p^2 −pq+q^2 +39=0 no real solution (2) x≥0∧y≥0 [⇒ 0≤x, y≤39] Let p, q ≥0∧x=p^2 ∧y=q^2 ⇒ p^2 +pq+q^2 −39=0 ⇒ q=−((p±(√(3(52−p^2 ))))/2) ⇔ y=39−(x/2)−((√(3x(52−x)))/2) [the “+” solution leads to y>39] Following pairs (x, y) and (y, x) are possible: (0, 39) (3, 27) (4, 25) (13, 13)

xy⩾0⇒x⩾0∧y⩾0∨x<0∧y<0(1)x<0∧y<0Letp,q>0∧x=−p2∧y=−q2⇒p2−pq+q2+39=0norealsolution(2)x⩾0∧y⩾0[⇒0⩽x,y⩽39]Letp,q⩾0∧x=p2∧y=q2⇒p2+pq+q2−39=0⇒q=−p±3(52−p2)2⇔y=39−x2−3x(52−x)2[the‘‘+″solutionleadstoy>39]Followingpairs(x,y)and(y,x)arepossible:(0,39)(3,27)(4,25)(13,13)

Commented by a.lgnaoui last updated on 08/Jan/23

thanks Solutions S={(13,13);(3,27);(27,3);(4,25);(25,4);(0,39)(39,0)}

thanksSolutionsS={(13,13);(3,27);(27,3);(4,25);(25,4);(0,39)(39,0)}

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