The n-th pentagonal number is given by the formula p_n = \dfrac{n(3n-1)}{2}. Prove that there are infinitely many pentagonal numbers that can be written as a sum of two perfect squares of positive integers.
Proposed by Jose Hernandez Santiago.
Solution:
We have p_n=n^2+\dfrac{(n-1)n}{2}=n^2+T_{n-1}, where T_{n-1} is the (n-1)-th triangular number. So, it is sufficient to prove that there are infinitely many triangular numbers which are perfect squares. Suppose that \dfrac{n(n+1)}{2}=m^2, \qquad m \in \mathbb{N}. This equation is equivalent to (2n+1)^2-8m^2=1 and putting x=2n+1, y=2m we have the Pell's equation x^2-2y^2=1, which has infinitely many solutions x=P_{2k}+P_{2k-1}, y=P_{2k}, where P_k=\dfrac{(1+\sqrt{2})^k-(1-\sqrt{2})^k}{2\sqrt{2}} is the k-th Pell number. Therefore, there are infinitely many triangular numbers m=P_{2k}/2 which are perfect squares and we are done.
No comments:
Post a Comment