When Is The Next Centered Pentagonal Flag?

The 50 stars on the American flag form two rectangles because $50$ is twice a square, $2 \times 25$. With statehood proposed for Washington, D.C., a $51$-star flag could place a star in the center surrounded by concentric pentagons of $5, 10, 15$ and $20$ stars, a centered pentagonal number. So $N=50$ has two nice properties at once: $N$ is twice a square and $N+1$ is a centered pentagonal number. After $50$, what is the next integer $N$ with both properties?

The Riddler, FiveThirtyEight(original post)

Solution

We need $N = 2q^2$ and $N + 1 = \tfrac{5p^2 + 5p + 2}{2}$ for integers $p,q$. Eliminating $N$, $4q^2 + 2 = 5p^2 + 5p + 2$, so $4q^2 = 5p(p+1)$, and after completing the square ($x = 2p+1$) this becomes $$x^2 - 80\,y^2 = 1, \qquad x = 2p+1,\ \ q = 5y,$$ a Pell equation. Its fundamental solution is $(x,y) = (9,1)$, and the rest come from powers of $9 + \sqrt{80}$: $$(9,1) \to (161,18) \to (2889,323) \to \cdots.$$ The first solution gives $p=4$, $q=5$, $N = 50$. The next, $x=161$, gives $p = 80$, $q = 90$, so $$N = 2\cdot 90^2 = \boxed{16200},$$ with $N+1 = 16201 = \tfrac{5\cdot 80^2 + 5\cdot 80 + 2}{2}$. (The one after that is $5{,}216{,}450$.)

The computation

Scan the centered pentagonal numbers: for each $p$, form $N = \tfrac{5p^2+5p}{2}$ and keep those for which $N/2$ is a perfect square (so $N$ is twice a square).

from math import isqrt
found = []
for p in range(1, 2000):
    N = (5*p*p + 5*p) // 2                 # N+1 is centered pentagonal
    if (5*p*p + 5*p) % 2 == 0 and N % 2 == 0:
        q2 = N // 2
        if isqrt(q2)**2 == q2:             # N = 2 * (perfect square)
            found.append(N)
print(found[:3])                            # [50, 16200, 5216450]