Advanced P-1 (BOMC-2)

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sourav das
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Advanced P-1 (BOMC-2)

Unread post by sourav das » Sat Mar 31, 2012 6:25 pm

1. (a) Prove that the sum of the squares of $3, 4, 5,$ or $6$ consecutive integers
is not a perfect square.
(b) Give an example of $11$ consecutive positive integers the sum of whose
squares is a perfect square.
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sourav das
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Re: Advanced P-1 (BOMC-2)

Unread post by sourav das » Sat Mar 31, 2012 6:37 pm

Intuition says:
for (a)
choose mod $8$ (First guess), If doesn't work then something else...
for (b)
We need to make an equation and then solve, If I choose $n,n+1,n+2,....,n+10$, then the equation i get not so beautiful. Then why don't we fix $x$ the middle man (x=n+5). I think It makes the equation beautiful...
I think my intuitions contain a high level of hint. So first please, try to make your own intuitions first, then if you really get stuck, share your intuitions with us and get some intuitions from us...
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Niloy Da Fermat
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Re: Advanced P-1 (BOMC-2)

Unread post by Niloy Da Fermat » Sat Mar 31, 2012 6:40 pm

a
for $ 3,4,5 $ integers, i considered mode $ 3,4,5 $ respectively.for $ 6 $ integers, we get
$ 6(x^2+x+3)=y^2-1 $
implies $ 4 $ divides $ y^2-1 $
from which we derive $ 2x(x+1)+3=0 (mod4) $
implies $ 4 $ divides $ 3 $, a contradiction
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Niloy Da Fermat
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Re: Advanced P-1 (BOMC-2)

Unread post by Niloy Da Fermat » Sat Mar 31, 2012 6:56 pm

b.
so far i have made
i guess $ x=1 $ is the only solution.we get here $ 11(x^2+10)=y^2 $
modular analysis is not helping much.
it has yielded me so far
$ x=\pm 1(mod66) $ :!: :!:
note: i made calculation error in this modular equation
Last edited by Niloy Da Fermat on Sun Apr 01, 2012 2:25 am, edited 1 time in total.
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Re: Advanced P-1 (BOMC-2)

Unread post by sourav das » Sat Mar 31, 2012 7:15 pm

@ Niloy
Note that $n$ odd>6 and $n^2\equiv 1$ (mod 11). Now check , I don't think problem setters don't set any real big number (Instead of a small one like a two digit number )
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shehab ahmed
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Re: Advanced P-1 (BOMC-2)

Unread post by shehab ahmed » Sat Mar 31, 2012 9:57 pm

Intuition
Seeing the problem,first thing came into my mind is quadratic residue.So far I have been able to solve the first two cases.For the first case,take (mod 3).U will only have 2 know that 2 is not a quadratic residue (mod 3).For the second case,take mod 4.u will only have to know that 2 isn't a quadratic residue mod 4.
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shehab ahmed
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Re: Advanced P-1 (BOMC-2)

Unread post by shehab ahmed » Sat Mar 31, 2012 10:00 pm

For the third,taking (mod 4) kills it.because,among 5 consecutive integers,either 3 of them is odd or 2 of them is odd
Last edited by *Mahi* on Sat Mar 31, 2012 10:03 pm, edited 1 time in total.
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shehab ahmed
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Re: Advanced P-1 (BOMC-2)

Unread post by shehab ahmed » Sat Mar 31, 2012 10:04 pm

Also among 6 positive integers,three of them must be odd and the result follows
Last edited by *Mahi* on Sat Mar 31, 2012 10:08 pm, edited 1 time in total.
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shehab ahmed
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Re: Advanced P-1 (BOMC-2)

Unread post by shehab ahmed » Sun Apr 01, 2012 12:08 am

I found two such sequence of 11 consecutive integers.One of them has 23 as its median and the other has 43 as its median.

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Re: Advanced P-1 (BOMC-2)

Unread post by nafistiham » Sun Apr 01, 2012 1:32 am

for b
suppose, the median is $x$ so, the summation is $(x-5)^2+(x-4)^2+\cdot\cdot\cdot+x^2+\cdot\cdot\cdot+(x+4)^2+(x+5)^2$
which is $y^2$ so, we have to solve $\frac{y^2}{11}=(x^2+10)$
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