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Prove that if $F_n$ an $L_n$ are Fibonacci and Lucas numbers respectively, and $n>2$, then

$$F_{n-2}\times F_{n-1}\times F_{n+1}\times F_{n+2}-15$$ $$F_{n-2}\times F_{n-1}\times F_{n+1}\times F_{n+2}-8$$ $$F_{n-2}\times F_{n-1}\times F_{n+1}\times F_{n+2}-3$$ $$F_{n-2}\times F_{n-1}\times F_{n+1}\times F_{n+2}+5$$ and $$L_{n-2}\times L_{n-1}\times L_{n+1}\times L_{n+2}-11$$ $$L_{n-2}\times L_{n-1}\times L_{n+1}\times L_{n+2}+9$$ $$L_{n-2}\times L_{n-1}\times L_{n+1}\times L_{n+2}+16$$

are always composite numbers.

These statements showed up while I did some modest research on similar topics - and I never saw these statements anywhere else, so I wanted to share them with you.

I have the concept of the proof for all cases, and I believe there is some beauty in the proofs - also not all seven cases have the identical "root cause". However, I will not reveal the proof yet, since I would like to see what would be your fresh approach, perhaps your methods are better.

Hope I did not make any mistake.

Just as a slight warning (and a hint), some of the cases produce fairly exotic factorization:


$$F_{4}\times F_{5}\times F_{7}\times F_{8}-8=61\times67$$ $$F_{31}\times F_{32}\times F_{34}\times F_{35}-8=157\times1163\times68035391\times12422650078087$$


$$F_{6}\times F_{7}\times F_{9}\times F_{10}-3=439\times443$$ $$F_{30}\times F_{31}\times F_{33}\times F_{34}-3=103\times1129\times2473\times1699499\times46068253393$$


$$F_{8}\times F_{9}\times F_{11}\times F_{12}+5=2917\times3137$$ $$F_{18}\times F_{19}\times F_{21}\times F_{22}+5=4561\times10037\times45751697$$ $$F_{23}\times F_{24}\times F_{26}\times F_{27}+5=5209\times1080553\times5628900677$$ $$F_{108}\times F_{109}\times F_{111}\times F_{112}+5=1024021\times1258717\times337819589\times222106191941\times395853050958193001\times24703386450231187973\times3809331398416771058561$$


$$L_{10}\times L_{11}\times L_{13}\times L_{14}+16=103681\times103687$$ $$L_{25}\times L_{26}\times L_{28}\times L_{29}+16=3079\times94933\times2031961\times62650261$$


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  • $\begingroup$ And this time, I won't make the mistake of assuming $F_n$ is a Fermat number... $\endgroup$ – abiessu Jan 7 '15 at 0:50
  • $\begingroup$ @abiessu Yeah, I was thinking about putting Fibonacci in bold haha... btw, if you don't know the solution by now, you will be surprised how simple the proof is... $\endgroup$ – VividD Jan 7 '15 at 1:05
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For the Lucas one, with odd $n,$ if you add another $9$ to your number you get another square, actually a fourth power,

3  4      Fermat:  ( 1 , 3 , 16 );  ( 3 , 10 , 29 ); 
4  7    = 2^3 cdot 13  cdot 23
5  11      Fermat:  ( 1 , 3 , 121 );  ( 1 , 87 , 149 ); 
6  18    = 3^2 cdot 107  cdot 109
7  29      Fermat:  ( 1 , 3 , 841 ); 
8  47    = 2^3 cdot 7 cdot 79  cdot 1103
9  76      Fermat:  ( 1 , 3 , 5776 ); 
10  123    = 2^3 cdot 3^2 cdot 13 cdot 97  cdot 2521
11  199      Fermat:  ( 1 , 3 , 39601 ); 
12  322    = 103681 cdot 103687
13  521      Fermat:  ( 1 , 3 , 271441 ); 
14  843    = 2^3 cdot 3^2 cdot 83 cdot 1427  cdot 59221
15  1364      Fermat:  ( 1 , 3 , 1860496 ); 
16  2207    = 2^3 cdot 7 cdot 73 cdot 769 cdot 2383  cdot 3167
17  3571      Fermat:  ( 1 , 3 , 12752041 ); 
18  5778    = 3^2 cdot 13 cdot 103 cdot 8311  cdot 11128427
19  9349      Fermat:  ( 1 , 3 , 87403801 ); 
20  15127    = 2^3 cdot 23 cdot 241 cdot 6829 cdot 8377  cdot 20641
21  24476      Fermat:  ( 1 , 3 , 599074576 ); 
22  39603    = 2^3 cdot 3^2 cdot 2389 cdot 54709  cdot 261399601
23  64079      Fermat:  ( 1 , 3 , 4106118241 ); 
24  103682    = 7 cdot 13  cdot mbox{BIG} 
25  167761      Fermat:  ( 1 , 3 , 28143753121 ); 
26  271443    = 2^3 cdot 3^2  cdot mbox{BIG} 
27  439204      Fermat:  ( 1 , 3 , 192900153616 ); 
28  710647    = 2^3 cdot 23 cdot 167  cdot mbox{BIG} 
29  1149851      Fermat:  ( 1 , 3 , 1322157322201 ); 
30  1860498    = 3^2 cdot 107 cdot 14431  cdot mbox{BIG} 
31  3010349      Fermat:  ( 1 , 3 , 9062201101801 ); 
32  4870847    = 2^3 cdot 7 cdot 13 cdot 739  cdot mbox{BIG} 
33  7881196      Fermat:  ( 1 , 3 , 62113250390416 ); 
34  12752043    = 2^3 cdot 3^2 cdot 331 cdot 409 cdot 3271  cdot mbox{BIG} 
35  20633239      Fermat:  ( 1 , 3 , 425730551631121 ); 
36  33385282    = 6263 cdot 54049  cdot mbox{BIG} 
37  54018521      Fermat:  ( 1 , 3 , 2918000611027441 ); 
jagy@phobeusjunior:

Fourth powers of $4,11,29,76, 199,$ every other Lucas number

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  • $\begingroup$ Yes, yes, yes! You are very close to the key! $\endgroup$ – VividD Jan 7 '15 at 1:50
  • $\begingroup$ @VividD, this is enough for me. $\endgroup$ – Will Jagy Jan 7 '15 at 1:52
  • $\begingroup$ Ok, I like seeing people who use C++ and STL (I think you do if I am not mistaken), and not these toys like Wolfram Mathematica... ;) Thanks for coming, and thanks for the answer! $\endgroup$ – VividD Jan 7 '15 at 1:57
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    $\begingroup$ @VividD Mathematica is far from a toy; oftentimes it's the best tool by far for the job. I trust its factoring routines much more than I trust my own, and its primality routines likewise, because they've been worked on extensively by multiple professionals in the field. $\endgroup$ – Steven Stadnicki Jan 7 '15 at 2:20
  • $\begingroup$ @StevenStadnicki, forget about "language wars"! Do you know the proof for all these seven expressions? I enjoyed our collaboration the other day... $\endgroup$ – VividD Jan 7 '15 at 2:28
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First three expressions are always composite since following identities are valid:

$$F_{n-2}F_{n-1}F_{n+1}F_{n+2} - 15 = ({F_n}^2 - 4)({F_n}^2 + 4)$$

$$F_{n-2}F_{n-1}F_{n+1}F_{n+2} - 8 = ({F_n}^2 - 3)({F_n}^2 + 3)$$

$$F_{n-2}F_{n-1}F_{n+1}F_{n+2} - 3 = ({F_n}^2 - 2)({F_n}^2 + 2)$$

They are all consequence of $x^2 - a^2 = (x+a)(x-a)$ and this identity:

$$F_{n-2}F_{n-1}F_{n+1}F_{n+2} + 1 = {F_n}^4$$

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