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Algebra

Blazing goIITian

 Joined: 2 Jun 2007 Post: 465
21 May 2009 21:44:30 IST
0 People liked this
7
725
good ones from numbers
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

1. find a prime divisor of integer n=4(3.7.11)-1 which is of the form 4k+3 wher k is any inetegr

2.find the integral solutions of x^4+y^4+z^4 -w^4=1995

plzz dont just give the answer solve properly really good ones from numbers

rates assured

more to come soon

thanking u

Blazing goIITian

Joined: 13 Aug 2008
Posts: 1313
21 May 2009 22:01:58 IST
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First one .... answer should be 71 ... (sorry had to manually sit n calculate this)

Second one .... good 1 from number theory ..... i guess you have to take modulo 16

x+ y4 + z4 = 1995 + w4

notice that if any no. n is even,  n4 is divisible by 16

when n is odd = 2m + 1 , then n2 is of the form 8k + 1 ; k is natural no.

so n4 == ( 8k + 1)2 = 64k+ 16k + 1 which on division by 16 yields remainder 1

So, N4 ==> 0 or 1  ( mod 16 )

But 1995 mod 16 = 11

So no integral solutions.

Blazing goIITian

Joined: 2 Jun 2007
Posts: 465
21 May 2009 22:11:20 IST
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second one okay but first one plzzz elaborate a bit

Blazing goIITian

Joined: 2 Jun 2007
Posts: 465
21 May 2009 22:16:09 IST
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seond one really good way of sloving but i cant get the first one and if u r done then post it theer are more to come for people who love numbers

Scorching goIITian

Joined: 10 May 2008
Posts: 289
22 May 2009 09:25:09 IST
2 people liked this

though this is not a rigorous proof, but a good way of ruling out options

we have to find a prime factor of the form 4k+3 which is same as 4p-1

prime numbers are also of the 6m+5 and 6m+1

CASE 1

we know  6m+1=4k+3 or 3m=2k+2 so m is a multiple of 2

putting m=2 in 6m+1 we can check 13 is the smallest prime factor. so other prime factors are  as n=923

from this case we see largest prime factor = 71

CASE 2

we know 6m+5 = 4p-1 or 3m+3=2p so p is a multiple of 3

putting p=3, we see 11 is not a divisor of 923

so next prime factor from this case can be 23 which after checking is not a factor

continuing this till p=18 (ie actually checking for 5 values)

we find smallest factor is 71. now one can check other factors should be

thus we conclude largest factor is 71

Forum Expert
Joined: 28 Feb 2007
Posts: 2185
22 May 2009 15:49:32 IST
2 people liked this

For the second one modulo 8 will work. Basically you have to chose a modulo in which there are as few residues as possible.

Since ,  and . You can easily see that the equation is inconsistent modulo 8

Blazing goIITian

Joined: 2 Jun 2007
Posts: 465
22 May 2009 22:03:13 IST
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thank u sir for ur answer i was actually thinking if u take modulo 16 then residue reamiing more so i wsa waiting for some other replies but u helped me as well as 2678 as well as rahul but folks plzz solve the first one with ur steps actually i ve forgotten the method of approaching these problems i had done them before

Forum Expert
Joined: 28 Feb 2007
Posts: 2185
23 May 2009 15:48:52 IST
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4*3*7*11 - 1 = 12*77 - 1 = 12 (6*12+5)-1

Put x = 12. Then the expression can be seen to be 6x2+5x-1 = (6x-1)(x+1) = 71*13

Its easy to see that 71 is the prime divisor in the sought form

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