Explaining a Fibonacci Explain why the number below is not 299th Fibonacci number:
222232244629420445529739893461909967206666939096499764990979600
I need an explanation
 A: Before spotting the easy argument given by WimC, I answered the question in a very different fashion. It’s ugly enough that I was going to ignore it, but now that I see that Jonah Sinick actually suggested it, I’ll toss it out for anyone who might be interested.
$F_{299}$ is the integer nearest to $$\frac1{\sqrt5}\left(\frac{1+\sqrt5}2\right)^{299}\;.$$
Let $n$ be your number. Then $n>2\times10^{62}$, so $\log_{10}n>62.3$. However, 
$$\log_{10}\frac1{\sqrt5}\left(\frac{1+\sqrt5}2\right)^{299}=299\Big(\log_{10}(1+\sqrt5)-\log_{10}2\Big)-\frac12\log_{10}5\approx62.1378\;,$$
and the difference between this and $62.3$ is too large to be attributable to roundoff error in the calculation with the logs. (With sufficient work one can justify that last claim rigorously.)
A: If you start with $1, 1, 2, 3, \dotsc$ then only every third Fibonacci number is even.  Now $299$ is not divisible by three.
A: I used GNP/PARI to find the answer using the formula $$F(n)=\frac{g^n-(-g)^{-n}}{\sqrt{5}}$$
where $g=(1+\sqrt{5})/2$.
$F(300)$ matches your result digit by digit.
A: There is a simple test whether a given number n is a member of the Fibonacci sequence:

*

*For Fibonacci numbers with an odd index, 5n²-4 must be a perfect square

*For Fibonacci numbers with an even index, 5n²+4 must be a perfect square

E.g. we have

*

*5*1²-4 = 1²

*5*1²+4 = 3²

*5*2²-4 = 4²

*5*3²+4 = 7²

*5*5²-4 = 11²

In the given case, the number passes the 5*n²+4 test case, so it is a Fibonacci number, but one with an even index, hence it can't be the 299th one.
In order to just disprove the 5 * n² - 4 case, we can e.g. work in modulo 7: The number is 4 mod 7, and 5 * 4² - 4 = 76 = 6 mod 7, but jacobi(6, 7) = -1, so this number can't be a square.
A: If $n\equiv 19\bmod 20$, then $F_n\equiv1\bmod  5$:
$1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,\color{blue}{1},0,1,1,2,...$
The given number misses the required residue.  It does, of course, hit the required residue $\bmod 5$ for $F_{300}$.
