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During my problem solving with Fibonacci numbers following thought crossed my mind.

How many Fibonacci numbers are there such that sum of its digits is a perfect square?

Here is a list of Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, and I observe that only 13 and 144 are satisfying the condition. Is there any other? How to analyze the situation?

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  • $\begingroup$ Did you forget 1? $\endgroup$ – TheRandomGuy Feb 28 '16 at 16:59
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In sequence A065411, the On-Line Encyclopedia of Integer Sequences has tabulated more Fibonacci numbers with this property. The next few are $806515533049393$, $61305790721611591$, $83621143489848422977$. There is probably not a good way to find these numbers other than just computing the sequence of Fibonacci numbers and testing whether the sum of digits is a square.

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  • $\begingroup$ Where are these in the sequence? Like, if $F_n$ equals the first number you've listed there, what's $n$? $\endgroup$ – Akiva Weinberger Feb 28 '16 at 5:06
  • $\begingroup$ Never mind: They're the 73rd, 82nd, and 97th Fibonacci numbers. $\endgroup$ – Akiva Weinberger Feb 28 '16 at 5:12

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