A Beautiful System from Moscow Mathematical Regatta

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[@mega_mango]

Респект от всех русских любителей математики!

[@hunterofinnovation007]

спасибо что делайте нам такой контенть

[@Nico2718_]

"so they're all equal" "oh it will be something pretty trivial now..." "so we have x(x²-x-1)" "ooooooooh"

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@mynamesgus4295]

what a lovely question

[@SyberMath]

Thank you! 😍

[@MortezaSabzian-db1sl]

Solving this device with three unknowns with the most up-to-date method in the world
1)x+y²=z³
2)x²+y³=z⁴
3)x³+y⁴=z⁵
We multiply the first equation by x and subtract it from the second equation and repeat the same thing for the second and third equations.
3) y²(y-x)=z³(z-x)
4)y³(y-x)=z⁴(z-x)
We multiply the right side of equation 3 by the left side of equation 4
y²(y-x)z⁴(z-x)=y³(y-x)z³(z-x)
Move all to the left and factor as much as possible
y²z³(y-x)(z-x)(z-y)=0
Substitute the results into the equations to get the answers

[@SyberMath]

Wow! Nice

[@nasrullahhusnan2289]

Note that:
• x, y, and x has similar coefficients but different power. Hence (x,y,z)=(0,0 0) is a solution.
• The power of z is sum of the power of x and y. Then (x,y,z)=(-1,-1,0)
• The power of z is 2 moreo than the power of x --> (x,y,z)=(±1,0,±1)
• Similarly with the power of y. Thus (x,y,z)=(0,1,1)

[@SyberMath]

Is that it?

[@richardl6751]

interesting how often the answer is the Golden Ratio.

[@vacuumcarexpo]

It's intriguing that the golden ratio appears.

[@SyberMath]

I know

[@hunterofinnovation007]

ураааааа россыйскые задачи ура

[@user-zy5tt8nu9h]

Za 🇷🇺👍

[@АпролЙцукенгович]

tooo much terrorusia. not cool dude