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

In this video, I'll be showing you step by step on how to solve this Olympiad Maths Algebra problem using a simple trick.
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[@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! 👏

[@shannonmcdonald7584]

7 and 5

[@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!

[@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.