A Very Nice Math Olympiad Problem | Solve for x and y | Algebra

[@key_board_x]

x² - y² = 24 → given: xy = 35 → y = 35/x x² - (35/x)² = 24 x² - (1225/x²) = 24 x⁴ - 1225 = 24x² x⁴ - 24x² - 1225 = 0 → let: X = x² ← where X ≥ 0 X² - 24X - 1225 = 0 Δ = (- 24)² - (4 * - 1225) = 576 + 4900 = 5476 = 74² X = (24 ± 74)/2 X = 12 ± 37 → we keep only the positive value (recall: X ≥ 0) X = 49 x² = 49 x = ± 7 First solution: x = 7 Recall: y = 35/x y = 5 Second solution: x = - 7 Recall: y = 35/x y = - 5

[@SpencersAcademy]

Fantabulous! 👏

[@azote2194]

(X^2 - Y^2) = (X+Y)(X-Y)=24; 24=12x2, the sum of X+Y and X-Y is 2X = 14 and X= 7 and Y= 5. This is the answer in natural numbers

[@biaohan4358]

The official solution to this kind of question should be to find x+y. Consider (x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=24^2+4*35^2=4*(144+1225)=4*1369, therefore x^2+y^2=74 or -74(if we solve in complex field. If we assume real solultions then the -74 can be discarded here). Then (x+y)^2=x^2+y^2+2xy= either 74+2*35=144 or -74+2*35=-4. Therefore x+y=12 or -12 or 2i or -2i, therefore x and y are the roots of the following equations: t^2-12t+35=0 or t^2+12t+35=0 or t^2+2i t+35=0 or t^2-2i t+35=0. It's easy to find x,y = 5 and 7 or x, y =-5 and -7 or x, y=7i, -5i or x, y=-7i, 5i. Plug in to find the 4 sets of solutions x=7, y=5 and x=-7, y=-5 and x=5i, y=-7i and x=-5i, y=7i.

[@eknathraykar3572]

A simple approach -- xy=35=7×5=(-7)×(-5). x,y must be both positive or both negative..x=+7, -7. y=+5, -5. These values satisfy both equations.

[@SpencersAcademy]

Nice one! 👏

[@aidandavis5550]

It didnt say the answers would be integers so there's no obvious reason to explore that factorization. that's one of those after-the-fact real analysis proofs that you probably never would have thought of until after you knew the answer.

[@eknathraykar3572]

@@aidandavis5550 I suggested a route. Out of the factors 1,5,7,35, choose those which satisfy the equation x^2 - y^2 = (x+y)(x-y) = 24. Again factorise 24, & realise that factors 12 and 2 are available here as (7+5) and (7-5). Seeking a solution by factorisation is a shorter and mathematicaly accepted route.

[@prollysine]

y^4-+24y^2-1225=0 , (y-5)(y^3+5y^2+49y+245)=0 , y=5 , y^3+5y^2+49y+245=0 , (y+5)(y^2+49)=0 , y= -5 , y^2= -49 , y= -5i , 5i , x=35/y , result (x , y) , (7 , 5) , (-7 , -5) , (-5i , 7i) , (5i ,-7i) , all test OK ,

[@SpencersAcademy]

Wonderful! 👏

[@prollysine]

@@SpencersAcademy Thanks!

[@shannonmcdonald7584]

7 and 5

[@biaohan4358]

You are missing the negative set, -7 and -5.