Solutions to x^3+...+y^3=y^3 in Integers

[anemone]

Find all solutions in integers of the equation

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

 

[Opalg]

[sp]$x^3+(x+1)^3+(x+2)^3+\cdots+(x+7)^3=8x^3 + 84x^2 + 420x + 784 = (2x+7)(4x^2 + 28x + 112) = z(z^2+63)$, 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 $(-5,-6),\: (-4,-4),\: (-3,4),\:(-2,6).$[/sp]

 

[anemone]

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 $(-5, -6)$.

 

[Opalg]

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.

 

[CellCoree]

The equation x^3+(x+1)^3+(x+2)^3+\cdots+(x+7)^3=y^3 represents a Diophantine equation, which is a type of mathematical equation that involves finding integer solutions. In this case, we are looking for solutions in the form of integers for both x and y.

To find all solutions, we can approach this problem using various methods such as brute force, algebraic manipulation, or number theory techniques. One possible method is to start with a specific value for x and find the corresponding value for y, then continue to iterate through different values of x until all solutions have been found.

Another approach is to use number theory techniques, such as the Pythagorean triples method, to find solutions. This method involves finding three integers, a, b, and c, such that a^2+b^2=c^2. In this case, we can rewrite the equation as (x+3)^3+(x+4)^3+(x+5)^3=(x+6)^3, which can be solved using the Pythagorean triples method.

It is important to note that this equation has an infinite number of solutions, as there are an infinite number of combinations of integers that can satisfy the equation. Therefore, it is not possible to list all solutions, but rather to provide a method for finding them.

In conclusion, finding solutions to x^3+...+y^3=y^3 in integers requires a systematic approach and may involve various mathematical techniques. It is an interesting problem that can be explored further using different methods and can lead to a deeper understanding of number theory and Diophantine equations.