(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.
:afrog:
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(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.
:afrog:
using the TI83 solver function:
if x=1, y=2, z=3 then w=-1.49
not sure how it does it though.
And if I ask for natural numbers?
And btw, that's not entirely accurate. If you take all of them on the LHS, you get a difference of -0.010023.
(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))
When w = -1.49Quote:
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.
x=(a+b)^.662252 - 1
y=(-b)^.662252 - 1
z=(a)^.662252 -1
a+b >= 0
a >= 0
b =< 0
w = -1
x = -(sqr(z^2+2*z-y*(y+2))+1) or
x = sqr(z^2+2z-y(y+2)) - 1
w <> 0
JFYI, this equation is known as Fermat's Last Theorem.
The web page shows the original equation.
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,b,c,d,e = 0Quote:
Originally posted by sql_lall
More interesting is more complex qs like this, for example:
a5+b5+c5=d5+e5
etc...
:confused:
a=(e^5+d^5-c^5-b^5)^(1/5)*(-(sqr(5)+1)/4 + sqr(-2*(sqr(5)-5))/4 * i)Quote:
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...
For the original:
x = y = z = w = infinity
Now tell me infinity is not a natural number :)
1) OK. Infinity is not a natural number. :D:D
(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.
(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
:lol:
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.
It's, unfortunately, been proven there is no natural number solution to that problem (and others like it). It is a cool equation, nonetheless.