Germany - Math Olympiad Question

A nice system of equations from math olympiad. What do you think about this question? If you're reading this ❤️.
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[@zawatsky]

Если (х+у)²=xy означает, что неполный квадрат суммы равен нулю. Таким образом x³-y³=0, при этом мы точно знаем про квадрат, а значит, x³=y³, т. е. х=у (нечётная степень). Пробуем подставить в неполный квадрат: x²+x²+x²=0, т. е. 3x²=0⇔х=0, а значит и у тоже. Но тогда сумма двух нулей получается равной восьми. Выходит, что решений (во всяком случае, действительных) не имеется.

[@jan-willemkrans7074]

Always fun to see seemingly simple statements to boil down to complex solutions. Good exercise, but this one wasn't too hard to solve for me and only took a few steps to complete. (Note that I used the textbook complete the square method for this as there's no leading coefficient for x or y squared after the first step. After that, you almost immediately get the answers.)
Perhaps people would appreciate the logic behind why complex solutions appear here. It's in the maximizing and minimizing values for the addition, then multiplication of the system. A multiplication of values with a hard limit on the sum like in this system of equations is always maximized with real numbers if the multiplied values are equal. From the first equations you would then get 4+4=8, but in the second 4*4=16, so it doesn't match. To get to 64 it simply has to come from a higher dimension. A similar thing happens with similar simple systems of equations that contain a complex number on one side each. You then sometimes wind up solving them with another higher dimension, too. That's in the form of quaternions (and onward number systems, but these subjects aren't taught much in universities worldwide).

[@ShineMultiDevice-zi1iu]

I use the same solution with searching some variable value, i try search the value of y first but, fortunately i still get the same answer with bit of Missing Point 😅

[@James-bak5]

I saw it and i thought well thats a quadratic and i got it after using the formulae but i can also understand the double equation method

[@Myroslav1911]

Using AM-GM inequality we can say that there is no real solution

[@RubelSk-m9r]

Give that,
x+y = 8 ------>(I)
xy = 64
(x-y)² = (x+y)² -4xy
(x-y)² = 8² - 4×64
(x-y)² = -3×64
(x-y) = √(-3×64)
(x-y) = 8√3i. ----->(ii)
From (i) & (ii),
2x = 8√3i + 8
x = 4(√3i + i)
y = 8 - x
y = 8 - 4(√3i + 1)
y = 4 - 4√3i
y = -4(√3i-1)

[@HangingQueen]

No solution

[@Sham_Fl]

Imaginary numbers

[@Christopher-e7o]

X,2×+5=8

[@NounPhechnai]

1st