Question

Question #0d004

Answer 1

The question seems to be wrong.

Explanation:

`a+b+c=0` lets check the value of `a^2017+ b^2017 +c^2017` for 2 sets of solutions.
If a=b=c=0 then,`a^2017+ b^2017 +c^2017` = 0 but when `a=1,b=1,c=-2` then `a^2017+ b^2017 +c^2017` = `1+1+(-2)^2017` ≠ 0

Answer 2

There are infinitely many answers.

Explanation:

If `a + b + c = 0`, then the addition of the variables has to add to zero.

Since you didn't say that `a^2017 + b^2017 + c^2017 = 0`, I can't argue that each variable could be `0`. However, I could say that

`a^2017 + b^2017 + c^2017 = x`

which can also be written as

`a^2017 + b^2017 + c^2017 - x = 0`

which would give a lot more leeway in an answer, because then you're just working for a constant `x`.

One answer (because there are now infinitely many) is if

`a = -1`
`b = -1`
`c = 2`
`c^2017 ~~ 1.5 * 10^607` ^
`x ~~ 1.5 * 10^607` ^^

then

`a + b + c = 0`

`-1 + (-1) + 2 = 0`

From there, you have

`(-1)^2017 + (-1)^2017 + 2^2017 - (1.5 * 10^607)`

The one problem here is that, of course, `(-1)^2017 = -1`. So, when using basic approximations with such large numbers, you'd still get `-2` as an answer, unless of course you were exactly precise. Wolfram Alpha (see notes below) tells me that there are 608 decimal places in `2^2017`, so you'd be doing a lot of work for not a lot of payoff.

The point here, however, is that there is an answer to this problem--in fact, there's infinitely many--but it's much too complex for anyone to do by hand using basic algebra. There might be a way to do this with calculus and limits, though I will freely admit I'm not sure what that would be.

Notes:

^: See https://www.wolframalpha.com/input/?i=2%5E2017 for the scientific notation of `2^2017`.
^^: See https://www.wolframalpha.com/input/?i=2%5E2017+%2B+2 for the scientific notation of `2^2017 + 2`. Hint: It's practically the same number at that point.

Edit: Of course, it just occurred to me that with the equation of `a^2017 + b^2017 + c^2017 - x = 0`, you could hypothetically just say that each variable equals `0` and any string of zeroes added/subtracted together will yield `0`, which shortens this answer considerably.