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

Sunday, October 30, 2011

Please email your solutions to puzzles@cotpi.com.

### 2 comments

### Credit

This is an original puzzle from cotpi.

### Further reading

The following is a list of resources on related topics:

- Weisstein, Eric W. "Fermat's Last Theorem." Wolfram MathWorld. 3 Nov. 2011. Wolfram Research. 6 Nov. 2011 <http://mathworld.wolfram.com/FermatsLastTheorem.html>.

## Philippe C. solved this puzzle:

The only solution is:

20

^{3}= 60 × (±10)^{2}+ 2000Let us prove it by solving 60a

^{2}+ 2000 = b^{3}.So, b

^{3}is a multiple of 20. Hence, b is a multiple of 10. Let b = 10c for some integer c.60a

^{2}+ 2000 = 1000c^{3}⇔ 3a

^{2}+ 100 = 50c^{3}So, a

^{2}is a multiple of 50. Hence, a is a multiple of 10. Let a = 10u for some integer u.300u

^{2}+ 100 = 50c^{3}⇔ 6u

^{2}+ 2 = c^{3}So, c is even. Let c = 2y for some integer y.

6u

^{2}+ 2 = 8y^{3}⇔ 3u

^{2}+ 1 = 4y^{3}So, u is odd. Let u = 2x + 1 for some integer x.

3(2x + 1)

^{2}+ 1 = 4y^{3}⇔ 3x

^{2}+ 3x + 1 = y^{3}⇔ (x + 1)

^{3}− x^{3}= y^{3}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.

x

^{3}= 60y^{2}+ 2000The right hand side is of the form 6a

^{2}b + 2b^{3}where a = y and b = 10. This can be rewritten as:6a

^{2}b + 2b^{3}= (a + b)^{3}− (a − b)^{3}Using this observation, we can now solve the equation as follows.

x

^{3}= 60y^{2}+ 2000⇔ x

^{3}= (y + 10)^{3}− (y − 10)^{3}⇔ x

^{3}+ (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 20

^{3}= 8000.