Japanese | Can you solve this ? | Math Olympiad

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[@davepelli3221]

At start you have to Say, Solving on integer numbers

[@christianlopez1148]

exactly, there are infinite solutions

[@johnlv12]

x-y is not always less than x x+x y +y y. Let x=1 and y=-1. Then, x-y=2 but x x+x y +y y=1-1+1=1

[@arunsanghvi6139]

Observe that 91 is sum of cubes of 4 and 3 I.e 91= 64 + 27. Hence, answer is x = 4 and y = -3

[@kimba381]

At no time did you indicate that x and y are integers. Your solution is incomplete until you check for other real, and indeed, complex solutions.

[@ccudmore]

45.5 x 2 also = 91
There’s actually an infinite number of solutions. This probably would map as an interesting 3d object

[@francescorusso7730]

​@@ccudmore
Two dimensional instead.

[@joeschmo622]

I just made a list of cubes of 1 to 10, ie, 1 to 1000.
Cubed x must be greater than cubed y, so x>=5 (125 for x=5 vs 64 for x=4).
The difference between adjacent cubes (ie, smallest distance) must also be less than 91, so y

[@Норайр-у7т]

Очень уж длинное решение,так явно видно,что х=6 а у=5;
Отрицательных ответов не надо для куба-неверно !!!
6³-5³= 91
216-125=91
91=91

[@ПавелКосырев-щ3к]

100-9=91

[@wrt717oo]

What had to be solved in whole numbers?

[@cyruschang1904]

x^3 = 91 - y^3
x = the cube root of (91 - y^3)

[@MartinPerez-oz1nk]

THANKS PROFESOR !!!, VERY INTERESTING !!!!

[@learncommunolizer]

You are welcome! Thank you very much!!

[@neilmccafferty5886]

a cubic with 4 solutions. cool. (a double cubic i suppose)

[@uveshkh1013]

How you assume that product of x and y is positive in the beginning?

[@Max_Carabelt]

Sir, because x.y=30 it is not posible for -4,3 and -3,4?

[@cacamalapasa1508]

NONSENSE: one equation with two unknowns, you need as many equations as degrees of freedom.

[@MrArcan10]

as I understand the only integer solutions required

[@sudarshandutta2498]

X=6, Y=5.

[@haothe1099]

X^3 - Y^3 = 91

[@1fessor]

6³-5³=91

[@ghanshyamlalgujar9831]

X=5&y=4

[@haothe1099]

1x91=91

[@吉田泰久-g2z]

x-y

[@user-tetris]

多分誤りですね...反例も容易に見つけられてしまいますし...

[@user-tetris]

(x,y) ∈ R²
x-y < x²+xy+y²
⇔x²+xy+y²-x+y > 0 ···①
だから
f(x,y):=x²+xy+y²-x+y=0
のグラフを書いてみる。
すると明らかに①を満たさない（x,y）∈Z² が存在することがわかる。
またf(x,y)は楕円の方程式3x²+y²=2を反時計回りに45度だけ回転させた後、x軸方向に1,y軸方向に-1だけ平行移動したもの。

[@windyyw]

wrong. true is 4（x^2+xy+y^2）≥（x-y）^2，that is （x+y）^2≥0。so only （1，91）and （7，13）can be possible solution。

[@user-tetris]

@@windyyw なぜx-y>0といえるの？

[@Bertin-q3y]

(6;5) (5;6)

[@heniwatisetiono6995]

x=6, y=5

[@adgf1x]

x=6 and y=5.

[@dinh1cam36]

Không có điều kiện x,y là số nguyên nên suy ra cách giải này sai.
Đáp án bài này là vô số nghiệm

[@daddykhalil909]

You based your solution on assumptions and trial and error
That isn’t logical nor acceptable

[@hansvangiessen8395]

Die Japanner snap niet niet veel van. Oneindig veel oplossingen. Niet de eerste keer dat hij in de fout gaat.

[@The_Conspiracy_Analyst]

Use soroban

[@bdfu4321]

废话太多了

[@ЭдуардПлоткин-р3л]

Почти 19 минут расписывать то,что решается за 4 минуты? Вы издеваетесь над нами?

[@johndavidmyself8039]

It's called "teaching."
"Teaching" takes longer than the learned doing.
There's no indication you've solved this.
But feel free to be snarky and irrelevant.

[@heinrich.hitzinger]

Everything is trivial if you know the solution. However, that's not the point.