How many perfect cubes are there that are two thousand more than sixty times some perfect square?

[SOLVED]

2 comments

Philippe C. solved this puzzle:

The only solution is:

203 = 60 × (±10)2 + 2000

Let us prove it by solving 60a2 + 2000 = b3.

So, b3 is a multiple of 20. Hence, b is a multiple of 10. Let b = 10c for some integer c.

60a2 + 2000 = 1000c3
⇔ 3a2 + 100 = 50c3

So, a2 is a multiple of 50. Hence, a is a multiple of 10. Let a = 10u for some integer u.

300u2 + 100 = 50c3
⇔ 6u2 + 2 = c3

So, c is even. Let c = 2y for some integer y.

6u2 + 2 = 8y3
⇔ 3u2 + 1 = 4y3

So, u is odd. Let u = 2x + 1 for some integer x.

3(2x + 1)2 + 1 = 4y3
⇔ 3x2 + 3x + 1 = y3
⇔ (x + 1)3 − x3 = y3

The only integer solutions are x = 0, y = 1 and x = −1, y = 1. The above equation has no other solutions from Fermat's Last Theorem.

Susam Pal from cotpi added:

Here is a shorter solution. Consider the Diophantine equation we get from the problem.

x3 = 60y2 + 2000

The right hand side is of the form 6a2b + 2b3 where a = y and b = 10. This can be rewritten as:

6a2b + 2b3 = (a + b)3 − (a − b)3

Using this observation, we can now solve the equation as follows.

x3 = 60y2 + 2000
⇔ x3 = (y + 10)3 − (y − 10)3
⇔ x3 + (y − 10)3 = (y + 10)3

The only solutions to this equation are x = 20, y = ±10. There are no other solutions to this equation as per Fermat's Last Theorem. So, the only possible such perfect cube is 203 = 8000.

Credit

This is an original puzzle from cotpi.

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