Jim sent in this solution, using the ideas from our hints.

a (n) = 1 + 2 + \dots + n = \frac{n (n + 1)}{2}

b (n) = 1^{2} + 2^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}

c (n) = 1^{3} + 2^{3} + \dots + n^{3} = \frac{n^{2} (n + 1)^{2}}{4}

It's obvious that

c = a^{2}

, from this.

Also, $2 a = n^{2} + n$ , so, solving the quadratic (and using the fact that $n > 0$ ), we get $n = \frac{- 1 + \sqrt{1 + 8 a}}{2}$ .

Now substitute this for $n$ in $b$ , to get $b = \frac{n (n + 1) (2 n + 1)}{6} = \frac{a}{3} \times (2 n + 1) = \frac{a}{3} \times \sqrt{1 + 8 a}$ . So $3 b = a \sqrt{1 + 8 a}$ , so $9 b^{2} = a^{2} + 8 a^{3}$ .

Now we can combine these two expressions: $9 b^{2} = c + 8 c \sqrt{c}$ , so $8 c \sqrt{c} = 9 b^{2} - c$ , so $64 c^{3} = 81 b^{4} - 18 b^{2} c + c^{2}$ .

(It's easy to check that the expressions above in terms of $n$ do work in this!)