1. ## solve in ihtegers

x,y integers

x^3 +(x+1)^3 +(x+2)^3+....+(x+7)^3=y^3

2. Originally Posted by perash
x,y integers

x^3 +(x+1)^3 +(x+2)^3+....+(x+7)^3=y^3
There's quite a bit of work going on here! I would think that the simplest start to this would be to expand and simplify:
$x^3 + (x + 1)^3 + (x + 2)^3 +....+ (x + 7)^3 = y^3$
becomes (after a fair amount of work)
$8x^3 + 84x^2 + 420x + 784 = y^3$

$8x^3 + 84x^2 + 420x + (784 - y^3) = 0$

$x^3 + \left ( \frac{21}{2} \right ) x^2 + \left ( \frac{105}{2} \right ) x + 98 - \frac{y^3}{8} = 0$

For this to be true for x and y integers we must have that
$\left ( \frac{21}{2} \right ) x^2 + \left ( \frac{105}{2} \right ) x - \frac{y^3}{8}$
must be an integer. That might be a good starting point.

-Dan

3. Thats about where i got, but isn't it 21/2 x² (84/8).

4. Originally Posted by topsquark
There's quite a bit of work going on here! I would think that the simplest start to this would be to expand and simplify:
$x^3 + (x + 1)^3 + (x + 2)^3 +....+ (x + 7)^3 = y^3$
becomes (after a fair amount of work)
$8x^3 + 84x^2 + 420x + 784 = y^3$

$8x^3 + 84x^2 + 420x + (784 - y^3) = 0$

$x^3 + \left ( \frac{21}{2} \right ) x^2 + \left ( \frac{105}{2} \right ) x + 98 - \frac{y^3}{8} = 0$

For this to be true for x and y integers we must have that
$\left ( \frac{21}{2} \right ) x^2 + \left ( \frac{105}{2} \right ) x - \frac{y^3}{8}$
must be an integer. That might be a good starting point.

-Dan
The typos have been fixed now.

-Dan