3
$\begingroup$

Solve following Diophantine equations:

$1) \ a^3-a^2+8=b^2$

2) $a, \ b,\ c \in \mathbb{Z^+}$$$\frac{a^3}{(b+3)(c+3)} + \frac{b^3}{(c+3)(a+3)} + \frac{c^3}{(a+3)(b+3)} = 7$$

3) $a^3-8=b^2$

In Problem 2 I tried to use inequality, then I can 'limit' that: $25 \ge a+b+c$ and $a^3 + b^3 + c^3 \ge 112$

Please use elementary way to solve it, I haven't studied elliptic curve yet, thanks.

$\endgroup$
3
  • $\begingroup$ As a hint for Problem 1), try adding 4 to both sides of the equation and see what happens... $\endgroup$
    – Mike Bennett
    Feb 12, 2013 at 19:40
  • $\begingroup$ Sorry professor but could you tell me more detail,I am not good in diophantine equations ? I have tried add 4 to both sides before but it led to $(x+2)(x^2 - 3x + 6)=y^2 + 4$ I also have $12 | y, x \equiv 2 \ (mod\ 3)$ but seem didn't help. $\endgroup$
    – Xeing
    Feb 13, 2013 at 9:44
  • 1
    $\begingroup$ Next step : which primes can divide $y^2+4$? $\endgroup$
    – Mike Bennett
    Feb 13, 2013 at 20:15

1 Answer 1

Reset to default
3
$\begingroup$

(2) Extremely ugly solution.

You have

$$a^3(a+3)+b^3(b+3)+c^3(c+3)=7(a+3)(b+3)(c+3)$$

or

$$(x-3)^3x+(y-3)^3y+(z-3)^3z=7xyz$$

Since it is symmetric, we can look for the solutions where $x \geq y \geq z$.

It is easy to check that for $x>15$ we have $(x-3)^3>7x^2$. This shows that $4 \leq x \leq 14$. For each particular $x$ you get a simpler solution which can be solved the same way.

P.S. It probably also helps observing that modulo 3 you have

$$(w-3)^3w \equiv 0,1 \pmod 3$$

Then, if none of $x,y,z$ is divisible by $3$ the LHS is 0 $\pmod 3$ which is not possible.

If one of $x,y,z$ is divisible by $3$, then all of them must be divisible by 3, and looking at the equation it follows that one of them is divisible by 9.

This should solve the equation.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .