Ice cream quiz

by David Radcliffe

This problem was posted in an ice cream shop in Oxford, England. I found it surprisingly difficult.

ice cream quiz

Here is my solution. Suppose that x + 3 = a^3 and x^2 + 3 = b^3. After substituting,

a^6 > (a^3 - 3)^2 + 3 = b^3

which implies that a^2 - 1 \ge b. Therefore

(a^3 - 3)^2 + 3 \le (a^2 - 1)^3

a^6 - 6a^3 + 12 \le a^6 - 3a^4 + 3a^2 - 1

3a^4 - 6a^3 - 3a^2 \le -13

3a^2 (a^2 - 2a - 1) \le -13

But the left side is positive for a \ge 3, so there are no whole number solutions.

Image credit: Janet McKnight

Thanks to @samuelprime and @nodrogadog for pointing out mistakes in earlier versions of this post.

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