I did a quick test of the first 54 Fibonacci numbers. No such examples apart from the trivial one 1 times 2 = 2. It is known that there are only finitely many Fibonacci numbers that are prime powers and the only square Fibonacci numbers are 1 and 144. The result most similar to what you are looking for is the proof by Luo Ming in 'On triangular Fibonacci numbers' that there are only finitely many Fibonacci numbers that are triangular - perhaps that proof could be adapted?

@James Tuite: Just now I have read the paper by Luo Ming. He proved that 8 Fn + 1 is a perfect square if and only if n = -1, 1, 0, 2, 4, 8, 10. Here n^2+n=F_k, implies that n = [-1(+or-)sqrt(1+4F_K)]/2. Further n is integer means 1+4F_K must be a perfect square. Is thare any such results?

Hi! I'm afraid that I don't know of exactly such a result. I mention Ming's paper, since the triangular numbers have form (n^2+n)/2, so it is similar. Although I appreciate that a factor of two can make a big difference!

Step 1: Revisiting the Initial Assertion

The question posed inquires whether there exists a Fibonacci number F_k such that n^2 + n = F_k for some integer n other than 0 and 1.

Step 2: Understanding the Expression n^2 + n

This expression can be factored as n(n + 1), representing the product of two consecutive integers. This implies that one of the factors is even and the other odd, as any pair of consecutive numbers will always include one number of each parity.

Step 3: Observing the Nature of the Fibonacci Sequence

The Fibonacci sequence is defined by F_0 = 0, F_1 = 1, and F_k = F_(k-1) + F_(k-2) for k > 1. Each term is the sum of the two preceding ones. This creates a progression where, apart from the initial cases, the difference between terms increases as the sequence progresses.

Step 4: Analyzing the Possibility of F_k = n(n + 1)

For a Fibonacci number F_k to equal n(n + 1), F_k would need to be the product of two consecutive integers. However, after F_2 = 1 and F_3 = 2, Fibonacci numbers, being the sum of their two predecessors, tend to diverge further from this form of consecutive integers' product. This is because, as we advance in the Fibonacci sequence, the difference between any two consecutive terms grows, making it impossible for them to be expressed as the product of two consecutive numbers, except in the cases of F_0 = 0 and F_3 = 2.

Step 5: Concluding on the Possibility of Other n

Given the growth of the Fibonacci sequence and the nature of the expression n(n + 1), we can conclude that, beyond n = 0 (which leads to F_0 = 0) and n = 1 (correcting the previous reference, which would lead to F_3 = 2 if we consider F_1 = 1 and F_2 = 1), there are no values of n that satisfy the equality n^2 + n = F_k for k > 3. This conclusion is derived both from the analysis of the product form n(n + 1) and from the progression and properties of the Fibonacci sequence.