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Jotadiolyne Dicci
Mar 5, 2020
  ·  Edited: Mar 6, 2020

For which values of n, is this true ?

Hi guys, so i have a question for you ! 2n^2 + (8/3)n - (5/3) = a^2 where a is a positive integer For which values of n, is this true ?

Bonus : Let a,b,c be three integers (a is non-zero), for which values of n, this equation is true: an^2+bn+c=d^2 with d a positive integer.


Hope to you !

P.S. : n is a positive integer everytime

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Eric Steckx
Mar 6, 2020

n must be of the form 6k+4

With this, I found empirically the first four : 10, 34, 12274 and 39994, but not yet the general law.

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Eric Steckx
Mar 6, 2020

I throw in the towel.

The formula looks like that of Fibonacci numbers or rather that of Pell numbers (because of the square root of 2), but it is much more complicated and I cannot demonstrate it.

Does this formula give all the solutions or only a part of them?

Jotadiolyne Dicci
Mar 6, 2020

This formula don't give all the solutions, just a part of them. It's very complicated.

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Eric Steckx
Mar 6, 2020

In your bonus question, there is no "n" in the formula !

Jotadiolyne Dicci
Mar 6, 2020

Ok, i corrected it.

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Eric Steckx
Mar 7, 2020


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Jotadiolyne Dicci
Mar 6, 2020

Thanks but there is a problem. If k=2 then 6k+4=16. But 2(16)^2+(8/3)(16)-(5/3) = 553. The square root of 553 is not an integer. So, n is not in this form. However, these integers (10, 34, 12274 and 39994) are true.


I can add one more thing. The teacher gave me one formula where it works.

n = 1/12 (208 (17 - 12 sqrt(2))^(2k) - 147 sqrt(2) (17 - 12 sqrt(2))^(2k) + 208 (17 + 12 sqrt(2))^(2k) + 147 sqrt(2) (17 + 12*sqrt(2))^(2k)-8)


There are still a few more to find out according to him. First of all, i don't know how he finds this. Yet, if we understand how he did it maybe we could find the other formulas.

Eric Steckx
Mar 6, 2020

When I wrote "n must be of the form 6k+4", I would say that all the solutions are of this form, not that all the 6k+4 are solutions !

Searching with wolframalpha, I can write that the solution is not polynomial.

Your teacher's formula is a monster ! I keep searching.

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Jotadiolyne Dicci
Mar 6, 2020

Ooooo ! Ok sorry !

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Eric Steckx
Mar 6, 2020

I found a more classical form.

This is a Pell-Fermat equation that we can write t² - 18a² = 46, where the "a" is the "a" of the initial formula. "a" solved, it is easy to solve for n with the quadratic formula.

I try to solve this Diophantin equation tomorrow or sunday.

Jotadiolyne Dicci
Mar 7, 2020

I'm sorry, but there are some things I don't understand. Why did you multiply a^2 by 18 and why does the equation equal to 46 ? I didn't know about Pell's equation. At least it seems like a good way to find "a" and then "n."

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Eric Steckx
Mar 7, 2020

Ok ! I got the answer. Give me time to write it legibly (in a few hours)

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Eric Steckx
Mar 7, 2020

The solution of the bonus question involves the same method :

- multiply the whole expression by "a" to have a square as coefficient for n²

- replace n by (t/a-b/2a) to eliminate the term at the first power

- put the terms in the canonic form of the Pell's equation

- ... and solve. The major difficulty lies in finding the first couples (t,a). If it is not too big, use a spreadsheet.

I'm not a mathematician, so I could miss some theoretical subtleties, but I think that this is the general concept.

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Jotadiolyne Dicci
Mar 7, 2020

WOW ! Impressive ! Thanks !