The sum of three numbers is 4.

[Pumblechook]

The sum of three numbers is 4, the sum of their squares is 10 and the sum of their cubes is 22. What is the sum of their fourth powers?

 

[rock.freak667]

The sum of three numbers is 4, the sum of their squares is 10 and the sum of their cubes is 22. What is the sum of their fourth powers?

If the unknown numbers are a,b,c,d and

Your data can now be written as

a+b+c+d=4
a²+b²+c²+d²=10
a³+b³+c³+d³=22

how do you think you would have to get a⁴+b⁴+c⁴+d⁴?

 

[jacksonpeeble]

If the unknown numbers are a,b,c,d and

Your data can now be written as

a+b+c+d=4
a²+b²+c²+d²=10
a³+b³+c³+d³=22

how do you think you would have to get a⁴+b⁴+c⁴+d⁴?

Where did the d value come from?
Also, I see the pattern, but are you sure this works?

This sounds more like a riddle to me, so you could just think about it very carefully ;-). Might I suggest looking at negative numbers as possibilities? I didn't actually solve the problem, but that seems like it might work.

 

[rock.freak667]

Where did the d value come from?

my bad, I read it as the sum of 4 numbers...it should be

$$a+b+c=4$$

$$a^2+b^2+c^2=10$$

$$a^3+b^3+c^3=22$$

If one felt inclined, one could attempt to solve these 3 equations, with 3 unknowns and then calculate $a^4+b^4+c^4$, or consider a,b,c to be the roots of a cubic polynomial and then use the relations of the roots to the coefficients and go from there with some helpful sums.

 

[Count Iblis]

The cubic polynomial method suggested by rock.freak667
is indeed the way to go.

We define Sn to be the sum of the n-th powers:

a + b + c = S1,

a^2 + b^2 + c^2 = S2,

a^3 + b^3 + c^3 = S3,

We want to express

S4 = a^4 + b^4 + c^4

in terms of S1, S2 and S3.

Consider the polynomial

P(x) = (1 + ax)(1 + bx)(1 + cx)


If we take the Logarithm of this, we get:

Log[P(x)] = Log(1+ax) + Log(1+bx) + Log(1+cx)

(Log denotes natural logarithm)

Next, we use the series expansion:

Log(1+y) = y - y^2/2 + y^3/3 - y^4/4 + ...

to expand Log[P(x)]. The coefficient of x^n then becomes proportional to Sn:

Log[P(x)] = S1 x - S2/2 x^2 + S3/3 x^3 - S4/4 x^4 + ...


Now take the exponential function of both sides. The left hand sided then becomes:

exp{Log[P(x)]} = P(x)

On the right hand side we expand the exponential function using the series expansion

exp(y) = 1 + y + y^2/2 + y^3/6 + y^4/24 + ...

So, we have:

P(x) =

exp(S1 x - S2/2 x^2 + S3/3 x^3 - S4/4 x^4 + ...) =


1 + S1 x + (S1^2 - S2)/2 x^2 + (S1^3/6 - S1 S2/2 + S3/3) x^3 +

(S1^4/24 - S1^2 S2/4 + S1 S3/3 + S2^2/8 - S4/4 ) x^4 + ...


But since P(x) is a third degree polynomial, the coefficients of x^n for n >3 must be zero. If you equate the coefficient of x^4 to zero, you get:

S1^4/24 - S1^2 S2/4 + S1 S3/3 + S2^2/8 - S4/4 = 0 ------>

S4 = S1^4/6 - S1^2 S2 +4/3 S1 S3 + S2^2/2


If you substitute the numbers for S1, S2 and S3, you get S4 = 50.

If you are curious about the numbers a, b, and c, you can use that we now have the polynomial P(x) in expanded form:

P(x) = 1 + S1 x + (S1^2 - S2)/2 x^2 + (S1^3/6 - S1 S2/2 + S3/3) x^3 =

1 + 4 x + 3 x^2 - 2 x^3

Since P(x) was defined as

P(x) = (1 + ax) (1 + bx)(1 + cx),

the zeroes of P(x) are -1/a, -1/b, -1/c

Solving the polynomial then gives

2, 1 + sqrt(2) and 1 - sqrt(2)

for the numbers.