For which values of n, is this true ?

WOW ! Impressive ! Thanks !

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.

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.

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

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?

Ooooo ! Ok sorry !

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.

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.