# $x^3 + 1$ and Arithmetic Sequences

While thinking up interesting problems, I stumbled upon this interesting fact:

If $x = 3n^2,$ for some positive integer $n,$ then $x^3 + 1$ can be expressed as the product of three distinct positive integers that form an arithmetic sequence.

For instance, when $x = 12 = 3 \times 2^2,$ we have $12^3 + 1 = 7 \times 13 \times 19.$ When $x = 75 = 3 \times 5^2,$ we have $75^3 + 1 = 61 \times 76 \times 91.$

This statement is relatively easy to prove with some algebraic manipulation, but I'm having a lot of trouble with the converse. If $x^3 + 1$ can be expressed as a product of three positive integers that form an arithmetic sequence, then must $x = 3n^2$ for some positive integer $n$? If not, what are the counterexamples? I'd appreciate it if anyone here can provide some information about this! Note by Steven Yuan
1 year, 8 months ago

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If $x=3n^2$, then $x^3+1=\left(3n^2\right)^3+1=\left(3n^2-3n+1\right)\left(3n^2+1\right)\left(3n^2+3n+1\right)$

- 1 year, 8 months ago

Yeah, that’s how you prove the statement, but I’m asking about the converse.

- 1 year, 8 months ago

The converse is not true. Some counterexamples are:

$719^3 + 1 = 189 \times 1040 \times 1891$

$719^3 + 1 = 70 \times 1647 \times 3224$

$1004^3 + 1 = 67 \times 2765 \times 5463$

$1655^3 + 1 = 621 \times 2072 \times 3523$

$2736^3 + 1 = 1142 \times 1939 \times 2736$

and so on.

- 1 year, 8 months ago

Interesting. Is there an explicit formula or recursive definition to quantify all the counterexamples?

- 1 year, 8 months ago

Write $x^3+1 = (n-a)n(n+a) = n^3-a^2n.$ Write $x=n-b.$ Then we get the equation $3bn^2 - (a^2+3b^2)n+(b^3-1)=0.$ Your case is $b=1,$ which leads to $n = \frac{a^2}3 + 1$ and $x = \frac{a^2}3.$

But there are plenty of other solutions, e.g. $b=321,$ $a=851,$ $n=1040.$

In general, this equation describes a cubic surface (or, more accurately, an affine open subset of a cubic surface) on which we are looking for integral points. This seems like it's pretty hard in general...

- 1 year, 8 months ago

Well, let's try a different value for $x$. I will use the derivative of $x$, which is $dx/dn = 6n$. Now the value of $x^3 + 1$ has a different value for $x$.

Now let's put $2$ in for $n$ again. $x$ still remains equal to $12$, as $12 = 6 * 2$. Once again we have $12^3 + 1 = 7 * 13 * 19$.

However, putting $5$ in for $x$ upholds $x = 30 = 6 * 5$. Now we have a whole different scenario. The equation is now $30^3 + 1 = 27001$, which is only divisible by $1$. Therefore, the arithmetic sequence $a * b * c$ cannot output a value 27001.

I hope this helps. :)

- 1 year, 8 months ago

I'm not sure what you're doing here. From what I can tell, you defined the function $x(n) = 3n^2,$ took its derivative, then claimed that if $x'(n) = k,$ then $x(n) = k$ as well. However, that last part isn't true; $x'(n) = x(n)$ only for functions of the form $Ce^n.$ Can you clarify for me?

- 1 year, 8 months ago

You're asking if the value $x=3n^2$ for any positive integer $n$ has to be in order for $x^3 + 1$ to be a product of three integers. I'm showing that $x^3 + 1$ cannot be a product of three integer, such that A) $n$ is a positive integer, and B) $x$ is any equation other than $3n^2$. What I did was I assigned a different value for $x$ then did the math to see if $x^3 + 1$ can be a product of three integers. For some equations, this is not always the case.

This might not always be true, though. Try doing the same process, such that $x=an^2$ and $a$ is an integer other than 3.

- 1 year, 8 months ago

I have a problem $x^3+1=6$ then $x^3=5$ so x is not even an integer.Pls explain what do u mean by converse.

- 1 year, 5 months ago

The converse reverses the two parts of the conditional. So for the conditional "If $x = 3n^2$ for some positive integer $n$, then $x^3 + 1$ can be expressed as the product of three distinct positive integers that form an arithmetic sequence," its converse would be "If $x^3 + 1$ can be expressed as the product of three distinct positive integers that form an arithmetic sequence, then $x = 3n^2$ for some positive integer $n$."

- 1 year, 4 months ago

that may not be always true by the above example that i gave since 6=1X2X3 (1,2,3 is an AP).But x is not necessarily an integer.

- 1 year, 4 months ago

What about when $n=1$? What arithmetic sequence of integers multiplies to give us 28?

- 1 year, 8 months ago

$28 = 1 \times 4 \times 7$

- 1 year, 8 months ago