The sum of the first n perfect cubes is a perfect fourth power. If f(x) = xth such number n, find f(x).?

Question

Find a closed form solution for f(x) in terms of x.
To get started,
f(1) = 1
f(2) = 8, since 1³+2³+3³+4³+5³+6³+7³+8³ = 6^4
f(3) = 49, since ∑n³ = 35^4
etc...

Answer 1

The required function is:
f(n) = (1/4)*((3 + 2√2)^n + (3 - 2√2)^n - 2)

The sum of the first n perfect cubes is (n(n + 1)/2)², this to be a perfect fourth power needs n(n + 1)/2 to be a perfect square, so we must solve the Diophantine equation
n(n + 1) = 2z² in natural numbers n and z, or
4n(n + 1) + 1 = 8z² + 1, or
(2n + 1)² - 2(2z)² = 1, or if x = 2n + 1, y = 2z we obtain a Pell's equation /please follow the link below for details/:
x² - 2y² = 1
Its fundamental solution is x_{0} = 3, y_{0} = 2 and, according the general theory of Pell's equation, all its infinitely many solutions can be obtained by the formula:
x_{n} = ((x_{0} + y_{0}*√2)^n + (x_{0} - y_{0}*√2)^n)/2
/the expression for y_{n} we don't need/
Here are the first of them: (1,0), (3,2), (17,12), (99,70) etc.
In our case
x_{n} = ((3 + 2√2)^n + (3 - 2√2)^n)/2,
this is always odd, and back to n
f(n) = (x_{n} - 1)/2 what yields the required expression.

You can convince Yourself that for n = 1, 2, 3, 4, 5, 6 this produces the values in Lebowski's answer.

Answer 2

I can't find a function, although

f(1) = 1
f(2) = 8
f(3) = 49
f(4) = 288 since ∑n³ = 204^4
f(5) = 1681 since ∑n³ = 1189^4
f(6) = 9800 since ∑n³ = 6930^4