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

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.

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?

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

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

Ok, i corrected it.

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.

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.