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Johan_Gunardi
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 Factors of 7^(7^n) + 1   « on: Jun 15th, 2007, 5:26pm » Quote Modify Remove

Prove that for every nonnegative integer n, the number 7^(7^n) + 1 is the product of at least
2n + 3 (not necessarily distinct) primes.

//Title changed by Eigenray
 « Last Edit: Feb 21st, 2008, 7:36am by Eigenray » IP Logged
iyerkri
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 Re: Hard Problem!   « Reply #1 on: Jun 16th, 2007, 6:34pm » Quote Modify

 hidden: By induction. For n=0, 7^(7^n) + 1 = 8 = 2*2*2.   Assume for n. Let 7^(7^n) = a. Then, 7^(7^(n+1)) + 1 = a^7 + 1 = (a+1)(a^6 - a^5 + a^4 - a^3 + a^2 - a +1)   first term is product of 2n+3 prime terms. Second term is even, hence (power of 2)*(odd number). Clearly for n >0, the second term is greater than 4. Thus a^7 is a product of atleast 2n +3 + 2 = 2(n+1) + 3 primes (not necessarily distinct).
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Barukh
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 Re: Hard Problem!   « Reply #2 on: Jun 17th, 2007, 5:54am » Quote Modify

iyerkri, unfortunately, your solution doesn't work: the second term is odd, not even, so it remains to be shown it's not a prime number.
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iyerkri
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 Re: Hard Problem!   « Reply #3 on: Jun 17th, 2007, 1:28pm » Quote Modify

yes!, I should go back to school to learn division by two.
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Eigenray
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 Re: Hard Problem!   « Reply #4 on: Jul 29th, 2007, 5:35pm » Quote Modify

This is sort of cheating, but since a=77^n is 7 times a perfect square, we can do*

>Factor[(a^7 + 1)/(a + 1) /. a -> 7x^2]
(1-7x+21x^2-49x^3+147x^4-343x^5+343x^6) (1+7x+21x^2+49x^3+147x^4+343x^5+343x^6)

[How to discover this by hand, I'm not sure, but it follows from

1-a+a2-a3+a4-a5+a6 = (1+a)6 - 7a(1+a+a2)2]

Now, it's easily seen that both factors above are > 1, so the ratio (a7+1)/(a+1) is divisible by at least 2 primes, proving the result.

Followup: Show that (a7+1)/(a+1) is divisible by at least 2 distinct primes, which are different from all previous primes, and conclude that 77^n+1 is divisible by at least 2n+1 distinct primes.

*More generally, for p an odd prime, and a=p*x2, the polynomial (ap+1)/(a+1) factors over Z iff p=3 mod 4, and (ap-1)/(a-1) factors iff p=1 mod 4.  Why?  (Hint: this is actually related to this problem.)
 « Last Edit: Jul 29th, 2007, 7:25pm by Eigenray » IP Logged
Eigenray
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 Re: Hard Problem!   « Reply #5 on: Feb 21st, 2008, 7:45am » Quote Modify

on Jul 29th, 2007, 5:35pm, Eigenray wrote:
 Followup: Show that (a7+1)/(a+1) is divisible by at least 2 distinct primes, which are different from all previous primes, and conclude that 77^n+1 is divisible by at least 2n+1 distinct primes. ... For p an odd prime, and a=p*x2, the polynomial (ap+1)/(a+1) factors over Z iff p=3 mod 4, and (ap-1)/(a-1) factors iff p=1 mod 4.  Why?

Anybody want to try these?  E.g.,

f(x) = 1 + px2 + p2x4 + ... + pp-1x2p-2

is irreducible iff p=2 or p=3 mod 4 is prime.  (Hint: what (and more importantly, where) are the roots?)
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