I found this problem in a book for secondary students. It says it's from the Australian Mathematical Olympiad in 1982.(adsbygoogle = window.adsbygoogle || []).push({});

The sequence [tex]a_1,a_2,...[/tex] is defined as follows:

[tex]a_1 = 2[/tex] and for [tex]n \geq 2[/tex], [tex]a_n[/tex] is the largest prime divisor of [tex]a_1a_2...a_{n-1}+1[/tex].

Prove that 5 is not a member of this sequence.

I don't know if my analysis of the problem is correct, but if 5 were a member of the sequence, then [tex]a_1a_2...a_{n-1}+1=5q[/tex] for some [tex]q\in{\textbf{N}}[/tex].

For [tex]q[/tex] even, then [tex]a_1a_2...a_{n-1}+1[/tex] is even, which means [tex]a_1a_2...a_{n-1}[/tex] is odd. But [tex]a_1 = 2[/tex] so that [tex]a_1a_2...a_{n-1}[/tex] is even. A contradiction.

How do I show that for [tex]q[/tex] odd? Or do I have to resort to a different approach?

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# Prove 5 is not a member of the sequence

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