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- Feb 14, 2012
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Find all solutions in integers of the equation
\(\displaystyle x^3+(x+1)^3+(x+2)^3+\cdots+(x+7)^3=y^3\)
\(\displaystyle x^3+(x+1)^3+(x+2)^3+\cdots+(x+7)^3=y^3\)
Notice that if you put $x=-3$ in the formula for $f(x)$ then it becomes $(-3)^3 + (-2)^3 + (-1)^3 + 0^3 + 1^3 + 2^3 + 3^3 + 4^3$. All the negative terms cancel with positive terms and you are just left with $4^3$.Let \(\displaystyle f(x)=x^3+(x+1)^3+(x+2)^3+(x+3)^3+(x+4)^3+(x+5)^3+(x+6)^3+(x+7)^3=8x^3+84x^2+420x+784=y^3\)
$\vdots$
Checking all possibilities out lead us to the only 4 solutions, and they are $(-2, 6)$, $(-3, 4)$, $(-4, -4)$ and $(-5, -6)$.