Question

Find x, y, and z such that x³+y³+z³=k, for each k from 1 to 100.​

Answer 1

Answer:

Yes, as of September 6, 2019, this problem is completely solved. First of all, note that the equation is about integers (this is apparently overlooked in some of the answers). Second, a well known congruence argument shows:

There are no solutions whenever k=4,5mod9k=4,5mod9 .

This holds since it is not difficult to verify that for every integer nn , we have n3=0,±1mod9n3=0,±1mod9 .

Up till this year, there were just two values for kk in the range 1..1001..100 for which it was not known if there were a solution. These values are k=33,42k=33,42 .

Both values are now known to have solutions. See 42 is the answer to the question “what is (-80538738812075974)³ + 80435758145817515³ + 12602123297335631³?” for some background information on both values and see Chapter 4. n=x^3+y^3+z^3 for a slightly outdated list (apart from the two values above, also 7474 is missing from that list, but a solution is in the previous reference).

Answer 2

Information About The Question :-

Find x, y, and z such that x³+y³+z³=k, for each k from 1 to 100.

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$\huge\tt\fcolorbox{red}{bl}{Answer}$

Solution:

This Question can have lot of solutions as constraints are very less

there is no information whether x , y & z are integer

+ ve , Real numbers

k = 1

x= 1 , y = 0 , z = 0

x =0 , y = 1 , z = 0

x = 0 , y = 0 , = 1

k = 2

x= 1 , y = 1 , z = 0

x =0 , y = 1 , z = 1

x = 1 , y = 0 , = 1

k = 3

x= 1 , y = 1 , z = 3

x = 1 , y = 1 , z = 1

x = 1 , y = 0 , = 1

k = 4

x=∛3 , y = 1 , z = 0

x=∛2 , y = ∛2 , z = 0

x=∛2 , y = 1 , z = 1

This way we can have so many solution

Easiest :

x³+y³+z³=k,

x = ∛k , y = 0 , z = 0 will satisfy

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Hope It Helps You,

BY Genius Attitude Boy,

@ItxAttitude