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Title: Factors of 7^(7^n) + 1 Post by Johan_Gunardi on Jun 15th, 2007, 5:26pm 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 |
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Title: Re: Hard Problem! Post by iyerkri on Jun 16th, 2007, 6:34pm [hideb] 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). [/hideb] |
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Title: Re: Hard Problem! Post by Barukh on Jun 17th, 2007, 5:54am iyerkri, unfortunately, your solution doesn't work: [hide]the second term is odd, not even, so it remains to be shown it's not a prime number[/hide]. |
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Title: Re: Hard Problem! Post by iyerkri on Jun 17th, 2007, 1:28pm yes!, I should go back to school to learn division by two. :-[ |
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Title: Re: Hard Problem! Post by Eigenray on Jul 29th, 2007, 5:35pm This is sort of cheating, but since a=77^n is [hide]7 times a perfect square[/hide], we can do* >[hide]Factor[(a^7 + 1)/(a + 1) /. a -> 7x^2][/hide] [hide](1-7x+21x^2-49x^3+147x^4-343x^5+343x^6) (1+7x+21x^2+49x^3+147x^4+343x^5+343x^6)[/hide] [How to discover this by hand, I'm not sure, but it follows from [hide]1-a+a2-a3+a4-a5+a6 = (1+a)6 - 7a(1+a+a2)2[/hide] 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 [hide]a=p*x2[/hide], 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 [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_putnam;action=display;num=1169761594]this problem[/link].) |
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Title: Re: Hard Problem! Post by Eigenray on Feb 21st, 2008, 7:45am on 07/29/07 at 17:35:09, Eigenray wrote:
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: [hide]what (and more importantly, where) are the roots[/hide]?) |
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