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Solutions in Integers

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,706
Find all solutions in integers of the equation

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

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,712
$x^3+(x+1)^3+(x+2)^3+\cdots+(x+7)^3=8x^3 + 84x^2 + 420x + 784 = (2x+7)(4x^2 + 28x + 112) = z(z^2+63)$, where $z=2x+7$. Thus $z$ is an odd integer and $z(z^2+63) = y^3$. If $(y,z)$ is a solution then so is $(-y,-z)$, so concentrate on the case where $y$ and $z$ are both positive. Since $y^3 = z^3 + 63z$ it is clear that $y>z$ and so $y\geqslant z+1$. Therefore $z^3+63z \geqslant (z+1)^3 = z^3 + 3z^2 + 3z + 1$, so that $3z^2 -60z + 1 \leqslant 0.$ Writing this as $3z(z-20) + 1 \leqslant 0$, you see that $z<20$. When you check the odd numbers from $1$ to $19$, you find that the only solutions are when $z=1$ and $z=3$. Thus the solutions (including the negative ones) for $(y,z)$ are $(\pm4,\pm1)$ and $(\pm6,\pm3)$; and the solutions for $(x,y)$ are $(-5,-6),\: (-4,-4),\: (-3,4),\:(-2,6).$
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,706
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 $(-5, -6)$.
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,712
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 $(-5, -6)$.
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 $4^3$.

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.