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- Feb 14, 2012

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\(\displaystyle x^3+(x+1)^3+(x+2)^3+\cdots+(x+7)^3=y^3\)

- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

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\(\displaystyle x^3+(x+1)^3+(x+2)^3+\cdots+(x+7)^3=y^3\)

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- #2

- Feb 7, 2012

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- #3

- Feb 14, 2012

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Thanks** Opalg** for participating and you have gotten all 4 correct solutions!

Here is the method I saw somewhere that is different from that of Opalg and I wish to share it here...

Let \(\displaystyle f(x)=x^3+(x+1)^3+(x+2)^3+(x+3)^3+(x+4)^3+(x+5)^3+(x+6)^3+(x+7)^3=8x^3+84x^2+420x+784=y^3\)

Also

\(\displaystyle (2x+7)^3=8x^3+84x^2+294x+343\)

\(\displaystyle (2x+10)^3=8x^3+120x^2+600x+1000\)

If $x\ge 0$, we can say that \(\displaystyle (2x+7)^3<f(x)<(2x+10)^3\).

This implies \(\displaystyle (2x+7)<y<(2x+10)\) and therefore $y$ is $2x+8$ or $2x+9$. But neither of the equations

\(\displaystyle f(x)-(2x+8)^3=-12x^2+36x+272=0\)

\(\displaystyle f(x)-(2x+9)^3=-24x^2-66x+55=0\)

have integer roots.

So we can conclude that there is no solution with $x\ge 0$.

Notice also that if we replace $x$ by $-x-7$, we end up having \(\displaystyle f(-x-7)=-f(x)\). This means $(x, y)$ is a solution iff $(-x-7, -y)$ is a solution. Therefore, there are no solution with $x\le -7$. Therefore, for $(x, y)$ to be a solution, we must have $-6 \le x \le -1$.

Checking all possibilities out lead us to the only 4 solutions, and they are $(-2, 6)$, $(-3, 4)$, $(-4, -4)$ and $(-5, -6)$.

Here is the method I saw somewhere that is different from that of Opalg and I wish to share it here...

Let \(\displaystyle f(x)=x^3+(x+1)^3+(x+2)^3+(x+3)^3+(x+4)^3+(x+5)^3+(x+6)^3+(x+7)^3=8x^3+84x^2+420x+784=y^3\)

Also

\(\displaystyle (2x+7)^3=8x^3+84x^2+294x+343\)

\(\displaystyle (2x+10)^3=8x^3+120x^2+600x+1000\)

If $x\ge 0$, we can say that \(\displaystyle (2x+7)^3<f(x)<(2x+10)^3\).

This implies \(\displaystyle (2x+7)<y<(2x+10)\) and therefore $y$ is $2x+8$ or $2x+9$. But neither of the equations

\(\displaystyle f(x)-(2x+8)^3=-12x^2+36x+272=0\)

\(\displaystyle f(x)-(2x+9)^3=-24x^2-66x+55=0\)

have integer roots.

So we can conclude that there is no solution with $x\ge 0$.

Notice also that if we replace $x$ by $-x-7$, we end up having \(\displaystyle f(-x-7)=-f(x)\). This means $(x, y)$ is a solution iff $(-x-7, -y)$ is a solution. Therefore, there are no solution with $x\le -7$. Therefore, for $(x, y)$ to be a solution, we must have $-6 \le x \le -1$.

Checking all possibilities out lead us to the only 4 solutions, and they are $(-2, 6)$, $(-3, 4)$, $(-4, -4)$ and $(-5, -6)$.

Last edited:

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- #4

- Feb 7, 2012

- 2,785

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 $(-5, -6)$.

If you put $x=-2$ then you get $f(-2) = (-2)^3 + (-1)^3 + 0^3 + 1^3 + 2^3 + 3^3 + 4^3 + 5^3$. Again, the negative terms cancel with some of the positive ones, and the remaining terms illustrate the fact that $3^3 + 4^3 + 5^3 = 6^3.$

The other two solutions for $x$ work in a similar way, except that in these cases the negative terms outweigh the positive ones.