Prove 5 is not a member of the sequence

**guildmage** · Jun11-09, 10:14 PM

I found this problem in a book for secondary students. It says it's from the Australian Mathematical Olympiad in 1982.

The sequence

is defined as follows:

and for $LaTeX Code: n \\geq 2$ ,

is the largest prime divisor of $LaTeX Code: a_1a_2...a_{n-1}+1$ .

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 $LaTeX Code: a_1a_2...a_{n-1}+1=5q$ for some $LaTeX Code: q\\in{\\textbf{N}}$ .

For

even, then $LaTeX Code: a_1a_2...a_{n-1}+1$ is even, which means $LaTeX Code: a_1a_2...a_{n-1}$ is odd. But

so that $LaTeX Code: a_1a_2...a_{n-1}$ is even. A contradiction.

How do I show that for

odd? Or do I have to resort to a different approach?

**Dragonfall** · Jun11-09, 10:55 PM

Well, if that were the case, then

LaTeX Code: a_1...a_n+1=2^a3^b5^c=2*3*a_3...a_n+1

, so a=b=0. I can't go further. But maybe the chinese remainder theorem can be used, since $LaTeX Code: 7*a_4...a_n\\equiv 2*a_4...a_n\\equiv 4 (\\mod 5^k)$ for all k<c+1.

**ramsey2879** · Jun11-09, 10:59 PM

Originally Posted by guildmage

I found this problem in a book for secondary students. It says its from the Australian Mathematical Olympiad in 1982.

The sequence is defined as follows:

and for $LaTeX Code: n \\geq 2$ , is the largest prime divisor of $LaTeX Code: a_1a_2...a_{n-1}+1$ .

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 $LaTeX Code: a_1a_2...a_{n-1}+1=5q$ for some $LaTeX Code: q\\in{\\textbf{N}}$ .

For even, then $LaTeX Code: a_1a_2...a_{n-1}+1$ is even, which means $LaTeX Code: a_1a_2...a_{n-1}$ is odd. But so that $LaTeX Code: a_1a_2...a_{n-1}$ is even. A contradiction.

How do I show that for odd? Or do I have to resort to a different approach?

it can be shown that both 2 and 3 are not divisors since 2 and 3 are members of the sequence and the product + 1 is not divisible by prime factors of the product. If 5 was the largest prime factor then the number is a power of 5 and ends in 25. That would seem to call for a different approach.

**Tibarn** · Jun12-09, 12:48 AM

Here's where I started before I got lazy:

The first three terms of the sequence are 2, 3, and 7. It's easy to see that if p is in the sequence, then p does not divide $LaTeX Code: a_1a_2...a_{n-1}+1$ ; thus, a given prime can appear at most once. Now, if $LaTeX Code: a_1a_2...a_{n-1}+1$ has a prime factor other than 5, this cannot be 2 or 3 since they're in the sequence, so it would have to be a prime larger than 5. Therefore, the only way for the largest prime factor to be 5 is for $LaTeX Code: a_1a_2...a_{n-1}+1$ to be a power of 5.

**Vanadium 50** · Jun12-09, 08:11 AM

Originally Posted by Tibarn

Therefore, the only way for the largest prime factor to be 5 is for $LaTeX Code: a_1a_2...a_{n-1}+1$ to be a power of 5.

Aren't you done? Then $LaTeX Code: a_1a_2...a_{n-1}$ is divisible by 4. But there is at most one power of two in that product.

**guildmage** · Jun12-09, 11:56 AM

I see. Then I suppose it is not necessary to split the problem in to two cases (q even and q odd).