Solvable?

[mendhak]

(x+1)^w+3 + (y+1)^w+3 = (z+1)^w+3

Find any values for w, x, y, and z for which the above equation is satisfied.

[Acidic]

using the TI83 solver function:
if x=1, y=2, z=3 then w=-1.49
not sure how it does it though.

[mendhak]

And if I ask for natural numbers?

[mendhak]

And btw, that's not entirely accurate. If you take all of them on the LHS, you get a difference of -0.010023.

[DiGiTaIErRoR]

(x^3+3x^2+3x+1)(x+1)^w = -((y+1)^w*(y^3+3y^2+3y+1)-(z+1)^w*(z^3+3z^2+3z+1))

[DiGiTaIErRoR]

Originally posted by Acidic
using the TI83 solver function:
if x=1, y=2, z=3 then w=-1.49
not sure how it does it though.

When w = -1.49

x=(a+b)^.662252 - 1
y=(-b)^.662252 - 1
z=(a)^.662252 -1

a+b >= 0
a >= 0
b =< 0

[DiGiTaIErRoR]

w = -1
x = -(sqr(z^2+2*z-y*(y+2))+1) or
x = sqr(z^2+2z-y(y+2)) - 1

w <> 0

[mendhak]

JFYI, this equation is known as Fermat's Last Theorem.

The web page shows the original equation.

[sql_lall]

unfortunately, and very famously, there are no answers to your questions, as you ensure that all bases are >0 and all powers are >2.

More interesting is more complex qs like this, for example:
a⁵+b⁵+c⁵=d⁵+e⁵

etc...

[mendhak]

Originally posted by sql_lall


More interesting is more complex qs like this, for example:
a⁵+b⁵+c⁵=d⁵+e⁵

etc...

a,b,c,d,e = 0

[DiGiTaIErRoR]

Originally posted by sql_lall
unfortunately, and very famously, there are no answers to your questions, as you ensure that all bases are >0 and all powers are >2.

More interesting is more complex qs like this, for example:
a⁵+b⁵+c⁵=d⁵+e⁵

etc...

a=(e^5+d^5-c^5-b^5)^(1/5)*(-(sqr(5)+1)/4 + sqr(-2*(sqr(5)-5))/4 * i)

[jemidiah]

For the original:

x = y = z = w = infinity

Now tell me infinity is not a natural number

[sql_lall]

1) OK. Infinity is not a natural number.
(Can you count to infinity?) Infact, it isn't a number at all!

2) As with all these questions, it is looking for NON-TRIVIAL (as in, don't use all 1s or 0s) NATURAL (positive integer) solutions.



P.S. i believe "NATURAL" includes zero, but defined as positive integer above cos a solution with zero would be trivial.

[alkatran]

(x+1)^(w+3) + (y+1)^(w+3) = (z+1)^(w+3)

x=-1
y=-1
z=-1
w = whatever you want


this is the sort of problem you're almost better off deciding to simplify (x+1)^(w+3) to 0 in your head... but teachers don't like that very much

[fundu]

If the Natural number restriction is not there then the easiest solution is take w=-2

It will reduce the equation to x+y=z

And no need to tell much many different solutions U can have now.

Even if U take w=-3 or w=-1 there are many solution. As in the case of latter it is x^2 + y^2 = z^2

If the restriction is that w should be >= 0 makes the problem interesting. I will try 2 solve in next 2 days.

By the way I will say that I am enjoying the company of U all a lot.

[jemidiah]

It's, unfortunately, been proven there is no natural number solution to that problem (and others like it). It is a cool equation, nonetheless.