Find all Real Solutions

We find all real solutions x_{1}, ... x_{10} to the equation sqrt{x_{1}^{2} + ... + x_{10}^{2}} = cbrt{x_{1}^{3} + ... + x_{10}^{3}}. The solution involves some satisfying algebra.
00:00 First steps
00:44 Simultaneous equations
03:53 Positive and negative terms
06:26 Finding solutions

[@nathanisbored]

in other words, the original equation has no solutions except for the trivial case where everything disappears down to a monomial and you can just cancel out the radicals

[@JohannNeumann-ic3gy]

Nice! Another upload

[@lf506]

the final description of the structure of the solutions could be made slightly more concise using the kronecker delta, i.e. x_i = a d(i, j) = a if i == j, 0 else where j is the index of the non-zero term in the 10-tuple and i is free
excellent video

[@zathrasyes1287]

Very cool!

[@samueldeandrade8535]

It would be more elegant to make the difference of equations B - A, getting (a-x_i) in the terms, and simply noticing that such binomials must be greater than zero and so each term (x_i)²(a-x_i) must be zero.

[@Happy_Abe]

We could generalize this to any number of terms right?
Nothing here was specific to 10

[@matze9713]

I just checked for 11 terms. Still works.

[@Happy_Abe]

@@matze9713 yeah you can inductively do this for all n
I’m thinking of if this would work too for infinite series case
I believe it still would

[@adriamaychum1387]

I haven't done it or proved it, but I guess it should work cause this is actually a case of equations of the type ||x||_p = ||x||_q for x \in R^n and 1

[@DrBarker]

Yes, this will definitely work with the exact same argument for any finite number of terms x_1, ... , x_n.

[@Happy_Abe]

@@DrBarker the infinite series case too?

[@holyshit922]

Could you record video how to calculate the sum
sum(binomial(n,2k)*binomial(k,m),k=m..floor(n/2))
I have the correct result but guy who gave it to me doesn't want to share with me how he has got it
Result of sum proposed by me is s_{n,m} = (binomial(n-m,n-2m) + binomial(n-m-1,n-2m))*2^{n-2m-1}
but i don't know how to get this result

[@GeoffryGifari]

So the solution works for any nonnegative real number a?

[@samueldeandrade8535]

What do you mean? The solutions are 10-tuples of the form (0, 0, ... , a, ..., 0, 0), with "a" being a nonnegative real number ("a" can assume any position).

[@samueldeandrade8535]

In other words, the set of solutions is the nonnegative parts of the coordinate axes in R¹⁰.

[@GeoffryGifari]

@@samueldeandrade8535 Yes this answers it.

[@tomctutor]

Good in the Reals, now how about complex solutions!

[@DrBarker]

For example (1, -1, i, -i, 0, 0, 0, 0, 0, 0) works, and we should get a lot more possible solutions too. Not sure how we could find all of the complex solutions though, or if there's even a nice solution to this.

[@worldnotworld]

Don't understand the a>=0 restriction at the beginning. Square roots have two values, in this case one positive and one negative, no?

[@vangrails]

No, a square root is always bigger or equal to zero.

[@worldnotworld]

@@vangrails So the square root of -1 is...?

[@gametimewitharyan6665]

​@@worldnotworld For any positive real numbers x, there does exist two square roots of x which are +√x and -√x
But when we use the √x symbol, we mean the "principal square root" of x
This principal square root is DEFINED to always be the postive square root of the number x
This is convenient because we can multiply the principal square root of a number by -1 to get the negative square root of the number
You can read more about this on Maths Stackexchange or Wikipedia

[@gametimewitharyan6665]

​@@worldnotworld The √-1 is an imaginary number i, it does not lie on the real number line hence it is neither negative nor positive, it is imaginary

[@worldnotworld]

@@gametimewitharyan6665 That's right; it doesn't answer my question though: why is it asserted at the beginning that a must be positive or equal to zero?