Find all positive integers $n$ for which $1 + 5a_n.a_{n + 1}$ is a perfect square.

Solution 1:

The only such $n$ is $n=3$, with $$ 1 + 5 a_3 a_4 = 1 + 5 \cdot 70 \cdot 180 = 63001 = 251^2. $$ Let $b_n = a_n/10 = 2, 3, 7, 18, 47, \ldots$ for $n=1,2,3,4,5,\ldots$ . These are sums of consecutive odd-order Fibonacci numbers: $2 = 1+1$ (with the first $1$ being $F_{-1}$), $3 = 1+2$, $7 = 2+5$, $18 = 5+13$, $47 = 13+34$, etc. by induction. It soon follows that $b_n b_{n+1} = 5 F^2 + 1$ where $F$ is the Fibonacci number common to $b_n$ and $b_{n+1}$: $$ 2\cdot 3 = 5 \cdot 1^2 + 1,\phantom{M} 3\cdot 7 = 5 \cdot 2^2 + 1,\phantom{M} 7\cdot 18 = 5 \cdot 5^2 + 1,\phantom{M} 18\cdot 47 = 5 \cdot 13^2 + 1, $$ etc. So we're looking to make $$ 1 + 5 a_n a_{n+1} = 1 + 500 b_n b_{n+1} = 2500 F^2 + 501 $$ a square, and it's easy to see that $F = 5$ is the only positive integer that makes this happen even without the hypothesis that $F$ be a Fibonacci number. (For instance, if $2500 F^2 + 501 = y^2$ with $y>0$, we may factor $501 = y^2 - 2500F^2 = (y-50F) (y+50F)$, or bound $y$ between $50F$ and $50F+1$ once $F>5$, or use the technique I described in this Mathoverflow answer.) Therefore $n=3$ is the unique answer as claimed.