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
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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
x4 + 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 = 64k2 + 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.
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
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
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
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
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