x3+1x^3 + 1 and Arithmetic Sequences

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

If x=3n2,x = 3n^2, for some positive integer n,n, then x3+1x^3 + 1 can be expressed as the product of three distinct positive integers that form an arithmetic sequence.

For instance, when x=12=3×22,x = 12 = 3 \times 2^2, we have 123+1=7×13×19.12^3 + 1 = 7 \times 13 \times 19. When x=75=3×52,x = 75 = 3 \times 5^2, we have 753+1=61×76×91.75^3 + 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 x3+1x^3 + 1 can be expressed as a product of three positive integers that form an arithmetic sequence, then must x=3n2x = 3n^2 for some positive integer nn? 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=3n2x=3n^2, then x3+1=(3n2)3+1=(3n23n+1)(3n2+1)(3n2+3n+1)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 - 1 year, 8 months ago

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

Steven Yuan - 1 year, 8 months ago

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The converse is not true. Some counterexamples are:

7193+1=189×1040×1891719^3 + 1 = 189 \times 1040 \times 1891

7193+1=70×1647×3224719^3 + 1 = 70 \times 1647 \times 3224

10043+1=67×2765×54631004^3 + 1 = 67 \times 2765 \times 5463

16553+1=621×2072×35231655^3 + 1 = 621 \times 2072 \times 3523

27363+1=1142×1939×27362736^3 + 1 = 1142 \times 1939 \times 2736

and so on.

David Vreken - 1 year, 8 months ago

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

Steven Yuan - 1 year, 8 months ago

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Write x3+1=(na)n(n+a)=n3a2n.x^3+1 = (n-a)n(n+a) = n^3-a^2n. Write x=nb.x=n-b. Then we get the equation 3bn2(a2+3b2)n+(b31)=0. 3bn^2 - (a^2+3b^2)n+(b^3-1)=0. Your case is b=1,b=1, which leads to n=a23+1n = \frac{a^2}3 + 1 and x=a23.x = \frac{a^2}3.

But there are plenty of other solutions, e.g. b=321,b=321, a=851,a=851, n=1040.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 - 1 year, 8 months ago

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Well, let's try a different value for xx. I will use the derivative of xx, which is dx/dn=6ndx/dn = 6n. Now the value of x3+1x^3 + 1 has a different value for xx.

Now let's put 22 in for nn again. xx still remains equal to 1212, as 12=6212 = 6 * 2. Once again we have 123+1=7131912^3 + 1 = 7 * 13 * 19.

However, putting 55 in for xx upholds x=30=65x = 30 = 6 * 5. Now we have a whole different scenario. The equation is now 303+1=2700130^3 + 1 = 27001, which is only divisible by 11. Therefore, the arithmetic sequence abca * b * c cannot output a value 27001.

I hope this helps. :)

Devon Fair - 1 year, 8 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)=3n2,x(n) = 3n^2, took its derivative, then claimed that if x(n)=k,x'(n) = k, then x(n)=kx(n) = k as well. However, that last part isn't true; x(n)=x(n)x'(n) = x(n) only for functions of the form Cen.Ce^n. Can you clarify for me?

Steven Yuan - 1 year, 8 months ago

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You're asking if the value x=3n2x=3n^2 for any positive integer nn has to be in order for x3+1x^3 + 1 to be a product of three integers. I'm showing that x3+1x^3 + 1 cannot be a product of three integer, such that A) nn is a positive integer, and B) xx is any equation other than 3n23n^2. What I did was I assigned a different value for xx then did the math to see if x3+1x^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=an2x=an^2 and aa is an integer other than 3.

Devon Fair - 1 year, 8 months ago

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I have a problem x3+1=6x^3+1=6 then x3=5x^3=5 so x is not even an integer.Pls explain what do u mean by converse.

rajdeep brahma - 1 year, 5 months ago

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The converse reverses the two parts of the conditional. So for the conditional "If x=3n2x = 3n^2 for some positive integer nn, then x3+1x^3 + 1 can be expressed as the product of three distinct positive integers that form an arithmetic sequence," its converse would be "If x3+1x^3 + 1 can be expressed as the product of three distinct positive integers that form an arithmetic sequence, then x=3n2x = 3n^2 for some positive integer nn."

David Vreken - 1 year, 5 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.

rajdeep brahma - 1 year, 5 months ago

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

Blan Morrison - 1 year, 8 months ago

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28=1×4×728 = 1 \times 4 \times 7

Steven Yuan - 1 year, 8 months ago

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