If $$a_{n+3}=-a_{n+2}+2a_{n+1}+8a_n $$ for $$ a_0=a_1=a_2=1.$$ Then prove that $a_n$ is a perfect square.
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What have you tried? What tools do you know? What are your thoughts on the problem? – Steven Stadnicki Aug 03 '18 at 04:27
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An attempt of proof cannot rely solely on $a_{n+2}, a_{n+1}$ and $a_n$ being squares (and the recurrence). If $a_0 = 9$, $a_1 = a_2 = 25$, we get $a_3 = 97$. – Fimpellizzeri Aug 03 '18 at 04:46
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1The most relevant part in the answer of the duplicate post is to consider $a_n=b_n^2$, where $b_0 = +1$, $b_1=-1$, and $b_{n+2}=-b_{n+1}-2b_{n}$. – Math Lover Aug 03 '18 at 04:59
1 Answers
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trial
Thu Aug 2 21:26:39 PDT 2018
3 9 = 3^2
4 1 = 1
5 25 = 5^2
6 49 = 7^2
7 9 = 3^2
8 289 = 17^2
9 121 = 11^2
10 529 = 23^2
11 2025 = 3^4 5^2
12 1 = 1
13 8281 = 7^2 13^2
14 7921 = 89^2
15 8649 = 3^2 31^2
16 73441 = 271^2
17 7225 = 5^2 17^2
18 208849 = 457^2
19 393129 = 3^2 11^2 19^2
20 82369 = 7^2 41^2
21 2374681 = 23^2 67^2
22 935089 = 967^2
23 4473225 = 3^4 5^2 47^2
24 16394401 = 4049^2
25 32761 = 181^2
26 68541841 = 17^2 487^2
27 62678889 = 3^2 7^2 13^2 29^2
28 74666881 = 8641^2
29 599025625 = 5^4 11^2 89^2
30 51739249 = 7193^2
31 1743647049 = 3^2 31^2 449^2
32 3152036449 = 23^2 2441^2
33 749171641 = 101^2 271^2
34 19504077649 = 7^2 71^2 281^2
35 7210557225 = 3^6 5^2 17^2 37^2
36 37790971201 = 73^2 2663^2
37 132662764441 = 457^2 797^2
38 603635761 = 79^2 311^2
39 567049662729 = 3^2 11^2 19^2 1201^2
40 495459724321 = 409^2 1721^2
41 643468687225 = 5^2 7^2 13^2 41^2 43^2
42 4883848063249 = 257^2 8599^2
43 366767105769 = 3^2 23^2 67^2 131^2
44 14548678518529 = 17^2 89^2 2521^2
45 25255640199001 = 967^2 5197^2
46 6775853684209 = 2603047^2
47 160124854862025 = 3^4 5^2 31^2 47^2 193^2
48 55471974098401 = 7^4 97^2 1567^2
49 318984565099321 = 11^2 401^2 4049^2
50 1072958221993681 = 271^2 120871^2
Thu Aug 2 21:26:40 PDT 2018
Will Jagy
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