Wikipedia:Reference desk/Archives/Mathematics/2016 December 9

= December 9 =

Particular Diophantine equation
$$(q - p)x^2 - 2pxy + (p - q)x - p y^2 + py = 0$$, with $$p < q$$, $$p$$ and $$q$$ positive integers.

Can this be transformed into Pell's equation by some substitution? 24.255.17.182 (talk) 02:50, 9 December 2016 (UTC)


 * See Conic section. That will get it terms of x^2, y^2, and a constant, and you can check whether your particular parametrization gives integers in the canonical form. Loraof (talk) 16:18, 9 December 2016 (UTC)
 * I put z=x+y to get
 * q(x2-x)-p(z2-z)=0.
 * Then u=2x-1, v=2z-1 to get
 * qu2-pv2=q-p.
 * Finally putting t=u-v and s=qu-pv,
 * s2-pqt2=(q-p)2.
 * I think from there it depends on the factorization properties of Z[√pq] unless there's a reduction I'm not seeing. --RDBury (talk) 22:12, 9 December 2016 (UTC)


 * Pell's equation, according to our article, is a hyperbola in Cartesian coordinates. So I would think it's impossible to convert the OP's equation into an instance of Pell's equation if the OP's equation is a parabola or ellipse. This can be checked by using the eccentricity formula in terms of the parameters of the equation, given at Conic section. Loraof (talk) 03:11, 11 December 2016 (UTC)
 * I think there's a distinction here that perhaps not everyone will agree with. Pell's equation is given as a Diophantine equation, not just an equation. If it's just an equation and the problem is to find all solutions, or even all rational solutions, then it would be fairly trivial. But the fact that solutions are restricted to integers makes the method of solution and the structure of the set of solutions very different. So it's true that reducing the equation to the algebraic form of Pell's equation is fairly easy, but doing so while preserving the condition that the solutions must be integers is more difficult. So to me the meaning of Pell's equation has the integer solution requirement built in, otherwise it's just another hyperbola and not worth talking about. (I think the image given in the lead of the article is somewhat misleading in this sense. The caption says "six of the integer solutions", but all solutions are integers by definition.) --RDBury (talk) 10:18, 11 December 2016 (UTC)
 * Right -- my point was simply that being a hyperbola is a necessary condition to be converted into a Pell equation. Loraof (talk) 21:10, 11 December 2016 (UTC)

Diagonal sums in modified Pascal's Triangle
The formula relating Fibonacci numbers with diagonal sums of Pascal's triangle is
 * $$F_{n}=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor} \tbinom {n-k-1} k$$

(as illustrated here). Now suppose you first divide the rth row of the triangle by r, show the resulting sums are
 * $$\frac{L_{n-1}}{n-1}=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor} \tfrac{1}{n-k-1} \tbinom {n-k-1} k $$

where Ln is the nth Lucas number. For example 18/6 = 1/6 + 5/5 + 6/4 + 1/3 and 29/7 = 1/7 + 6/6 + 10/5 + 4/4. I think I'm converging on a proof but it's not pretty and it seems like an interesting formula. --RDBury (talk) 20:39, 9 December 2016 (UTC)


 * Looks like
 * $$\begin{align}

\sum_{k} \tfrac{1}{n-k-1} \tbinom {n-k-1} {k} & = \frac{1}{n - 1} \sum_{k} \tfrac{n - 1}{n-k-1} \tbinom {n-k-1} {k} \\ & = \frac{1}{n - 1} \left(\sum_{k}\tbinom {n-k-1} {k} + \sum_{k} \tfrac{k}{n-k-1} \tbinom {n-k-1} {k}\right) \\ & = \frac{1}{n - 1} \left(F_n + \sum_{k} \tbinom {n-k-2} {k - 1}\right) \\ & = \frac{1}{n - 1} \left(F_n + \sum_{k} \tbinom {(n - 2) - k - 1} {k}\right) \\ & = \frac{1}{n - 1} \left(F_n + F_{n - 2} \right) \end{align}$$
 * after you check that I haven't done any funny business with the bounds of summation. --JBL (talk) 18:07, 14 December 2016 (UTC)