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Topic: Quadratic divisibility (Read 448 times) |
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NickH
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Quadratic divisibility
« on: Jun 13th, 2003, 12:08pm » |
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Show that, if n is an integer, n2 + 11n + 2 is not divisible by 12769.
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ThudnBlunder
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Re: Quadratic divisibility
« Reply #1 on: Jun 13th, 2003, 4:31pm » |
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:Not very elegant, I know:
Assume n2 + 11n + 2 = 12769k = 1132k for some positive integer k
Solving for n, we get D2 = (113)[1 + 4*(113k)2]
The first term (113) is prime
and the 2nd term is one more than a pefect square.
Therefore D2 cannot be a perfect square.
Hence, n cannot be an integer.
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| « Last Edit: Jun 14th, 2003, 2:49am by ThudnBlunder » |
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Sir Col
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Re: Quadratic divisibility
« Reply #2 on: Jun 14th, 2003, 5:28am » |
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T&B, I found the disciminant to be 113+4*1132k?
Knowing NickH's problems, there's going to be some ingenious lateral step required. Here are my rambings so far...
By defining f(n)=n2+11n+2, it is clear that f(n) will always be even and it is possible to show that f(n)==+/-1 mod 3; that is, f(n) is a and not a multiple of 3.
Transferring this to the RHS, 12769k, we determine that k must be of the form 6a+2 or 6a-2 if a solution exists.
However, I've not been able to take this any further. Anybody got any other ideas?
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| « Last Edit: Jun 14th, 2003, 10:30am by Sir Col » |
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ThudnBlunder
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Re: Quadratic divisibility
« Reply #3 on: Jun 14th, 2003, 9:30am » |
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Quote:| T&B, I found the disciminant to be 17+4*1132k? |
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Duh. Yes, trying again I get 113*(1 + 4*113) = (113)(3)(151)
As these 3 numbers are all prime, D2 cannot be a perfect square.
But, as you say, this is surely not the expected method.
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Sir Col
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Re: Quadratic divisibility
« Reply #4 on: Jun 14th, 2003, 10:40am » |
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I just realised that I made a typo in my discriminant: I wrote 17 rather than 113?! 
I think you forgot k this time T&B.
However, as the discriminant is 113(1+4*113k), we can see that the second factor is one more than a multiple of 113; that is, we can write it as 113(113t+1). Clearly this cannot be square, as the second factor cannot contain a factor of 113.
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ThudnBlunder
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Re: Quadratic divisibility
« Reply #5 on: Jun 14th, 2003, 11:41am » |
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Quote:| I think you forgot k this time T&B. |
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Man, we are making hard work of this. 
Quote:| Clearly this cannot be square, as the second factor cannot contain a factor of 113. |
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I guess you mean '113 as a factor'.
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| « Last Edit: Jun 14th, 2003, 2:07pm by ThudnBlunder » |
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NickH
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Re: Quadratic divisibility
« Reply #6 on: Jun 22nd, 2003, 8:43am » |
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Consider n2 + 11n + 2 = (n + 62)2 - 113(n + 34).
If n2 + 11n + 2 is divisible by 1132, it must be divisible by 113, and so (n + 62) must be divisible by 113. But then (n + 62)2 is divisible by 1132 while 113(n + 34) is not!
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