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We have an arithmetic sequence:
Fixed LaTeX - Ricky
Let a(n-1)=x, a(n)=y.
By definition, a(n+1)=5x+3y
a(n+2)=5y+[3(5x+3y)]=5y+15x+9y=15x+10y
a(n+1)*a(n+2)=(5x+3y)(15x+10y)=45x² +30y² +95xy
It can be seen that a(n+1)*a(n+2) is not divisible by a(n), i.e.
y since two of the three terms forming the sum a(n+1)*a(n+2) are divisible by y, whereas the third term, that is, 45x² is independent of y.
Maybe my logic is wrong at some step, I tried making a start!
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
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Youve made a start but its not enough. 45x[sup]2[/sup] is not independent of y. If x is a[sub]n−1[/sub] and y is a[sub]n[/sub], then y = 5a[sub]n−2[/sub]+3x so they are related.
Last edited by JaneFairfax (2007-10-07 03:16:36)
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How does one do subscript without LaTeX?
There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.
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The standard ASCII method is an underline: a_n.
After banging on my head on this brick wall for two or three days, I think Ive finally found a solution. Im post the whole thing when I get home (Im now at a friends house).
First you establish two preliminary results:
P1: a[sub]n[/sub] is not divisible by 3 or 5 for any n.
P2: gcd(a[sub]n[/sub],a[sub]n+1[/sub]) = 1.
P1 is easily proved by induction, so Ill skip the proof here. P2 is also proved by induction; Ill post my proof later.
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So were all right for n = 1.
Hence, by induction, P2 is proved.
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Now were set. Suppose there is some n ≥ 2 for which a[sub]n[/sub] divides a[sub]n+1[/sub]a[sub]n+2[/sub]. Then
But gcd(a[sub]n[/sub],a[sub]n+1[/sub]) = 1. So a[sub]n[/sub] cant possibly divide a[sub]n+1[/sub][sup]2[/sup] unless a[sub]n[/sub] = 1 (which doesnt hold for n ≥ 2).
Contradiction! Quod erat demonstrandum.
Last edited by JaneFairfax (2007-10-08 10:35:34)
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