$x^3 + 1$ 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!

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Khang Nguyen Thanh - 3 years, 3 months ago

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

Steven Yuan - 3 years, 3 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.

David Vreken - 3 years, 3 months ago

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

Steven Yuan - 3 years, 3 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...

Patrick Corn - 3 years, 3 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. :)

Devon Fair - 3 years, 3 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?

Steven Yuan - 3 years, 3 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.

Devon Fair - 3 years, 3 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.

rajdeep brahma - 2 years, 12 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 $n$ ."

David Vreken - 2 years, 12 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.

rajdeep brahma - 2 years, 12 months ago

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

Blan Morrison - 3 years, 3 months ago

$28 = 1 \times 4 \times 7$

Steven Yuan - 3 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

x3+1x^3 + 1x3+1 and Arithmetic Sequences

Comments

$x^3 + 1$ and Arithmetic Sequences