Mexico, Math Olympiad: x^√y - y = 65; x, y =? x^√y > y > 65 or x^√y > 65 > y > 0, x, y ϵ R⁺; 65 = (1)(65) = (5)(13) x^√y - y = (1)(65) = (33 - 32)(33 + 32) = 33^2 - 32^2 = 33^(32/16) - 32^2 = (¹⁶√33)^√(32^2) - 32^2; x = ¹⁶√33, y = 32^2 x^√y - y = (5)(13) = (9 - 4)(9 + 4) = 9^2 - 4^2 = 3^4 - 16 = 3^√16 - 16; x = 3, y = 16 Answer check: x = ¹⁶√33, y = 32^2: x^√y - y = 65; Confirmed as shown x = 3, y = 16: x^√y - y = 65; Confirmed as shown Final answer: x = ¹⁶√33 = 1.244, y = 32^2 = 1024 or x = 3, y = 16 The calculation was achieved on a smartphone with a standard calculator app There are infinite solutions for this equation as @sy8246 has pointed out! The restrictions of “x, y ϵ R⁺” may be helpful in simplifying the math work.
Brilliant
Very nice! ❤
Thank you for explaining. ・・・ There are infinite solutions.
(x, y) = (66, 1), (√69, 4), (74^(1/3), 9), (3, 16), (90^(1/5), 25), ・・・ [and, (-√69, 4), (-3, 16), ・・・ ]
You are right! ❤
x、yに制限を付けないと与式を満たす組は無限にあることになる。ｘが負でも成り立つ組は無限にある。
あなたは正しいです、親愛なる ❤
Mexico, Math Olympiad: x^√y - y = 65; x, y =?
x^√y > y > 65 or x^√y > 65 > y > 0, x, y ϵ R⁺; 65 = (1)(65) = (5)(13)
x^√y - y = (1)(65) = (33 - 32)(33 + 32) = 33^2 - 32^2 = 33^(32/16) - 32^2
= (¹⁶√33)^√(32^2) - 32^2; x = ¹⁶√33, y = 32^2
x^√y - y = (5)(13) = (9 - 4)(9 + 4) = 9^2 - 4^2 = 3^4 - 16 = 3^√16 - 16; x = 3, y = 16
Answer check:
x = ¹⁶√33, y = 32^2: x^√y - y = 65; Confirmed as shown
x = 3, y = 16: x^√y - y = 65; Confirmed as shown
Final answer:
x = ¹⁶√33 = 1.244, y = 32^2 = 1024 or x = 3, y = 16
The calculation was achieved on a smartphone with a standard calculator app
There are infinite solutions for this equation as @sy8246 has pointed out!
The restrictions of “x, y ϵ R⁺” may be helpful in simplifying the math work.
Very nice trick! I really appreciate that ❤