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.
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