\(x^3 + 1\) 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
4 months ago

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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 - 4 months ago

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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 - 4 months ago

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

Steven Yuan - 4 months ago

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

Khang Nguyen Thanh - 4 months ago

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

Steven Yuan - 4 months ago

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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 - 1 month ago

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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 - 4 weeks ago

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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 - 3 weeks, 6 days ago

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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 - 4 months ago

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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 - 4 months ago

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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 - 4 months ago

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

Blan Morrison - 4 months ago

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\(28 = 1 \times 4 \times 7\)

Steven Yuan - 4 months ago

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