# $$x^3 + 1$$ 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
7 months, 1 week ago

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## Comments

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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...

- 7 months, 1 week 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.

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week 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.

- 4 months 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 $$n$$."

- 4 months 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.

- 4 months 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. :)

- 7 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?

- 7 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.

- 7 months ago

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

- 7 months, 1 week ago

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$$28 = 1 \times 4 \times 7$$

- 7 months, 1 week ago

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