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1.Find all solutions of the Ramanujan-Nagell diophantine equation x^2+7=2^n with x<=1000
2.Find all solutions of the Ljunggren diophantine equation x^2 - 2y^4=-1
3.For n>=1 consider the rational number, h(n)=1+1/2+1/3+......+1/n
Prove that h(n) is not an integer for any x>=2
4.Prove that n, n+2, n+4 are all primes if and only if n=3.
5.The prime numbers p and q are called twin primes if |p-q|=2, Prove that if p and q are twin primes greater than 3 then q+p is divisible by 12
6.Prove that if x,y,z are integer such that x^2+y^2=z^2 , then xyz=0(mod 60) [note that = is modula =]
This text book Elementary Methods in Number Theory is madness , some exercises even require you to prove theorems that put forward by Mathematician.Anyone read it ?written by Melvyn B.Nathanson.
Last edited by Stanley_Marsh (2007-02-15 06:36:52)
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First , I can solve half of the section , As, I learn more and more I can solve fewer and fewer problems...Help!
Here another two
7.Let 2=p(1)<p(2)<.... be the sequence of primes in increasing order Prove that p(n)<=2^2^(n-1) for all n>=1
8.Let a and b be positive integers with a>b . The length of the Euclidean algorithm for a and b, denoted by E(a,b) , is the number of divisions required to find the greatest common divisor of a and b . Prove that E(a,b)<= (logb/log@)+1 , where @= (1+5^1/2)/2 [I think this one is the hardest ]
Last edited by Stanley_Marsh (2007-02-15 06:52:24)
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#4. In another topic, it was shown that x, x+4 and x+8 can't all be primes. The same logic works for x, x+2 and x+4. Simply put, one of those 3 numbers has to be divisble by 3.
You won't ever have a pair of 4's in a row. Let's say X is a prime number. We know it's odd as they all are except for 2. Every odd number is going to be either:
A) divisible by 3 (x mod 3 = 0);
B) have a remainder of 1 when divided by 3 (x mod 3 = 1);
C) have a remainder of 2 when divided by 3 (x mod 3 = 2).Let's consider X, X+4 and X + 8 when X is prime:
A) X is divisible by 3 - can't happen when X is prime
B) X mod 3 = 1
Then (X+4) mod 3 = 2
and (X+8) mod 3 = 0.
Therefore X+8 is divisible by 3 and is not prime.
C) X mod 3 = 2
Then (X+4) mod 3 = 0
Therefore X+4 is divisible by 3 and is not prime.So for any prime number X, either X+4 or X+8 will be divisible by 3.
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5.The prime numbers p and q are called twin primes if |p-q|=2, Prove that if p and q are twin primes greater than 3 then q+p is divisible by 12
Logic similar to the that used to solve #4 can be used here. I don't know if you would call this a "proof" though. More of a demonstation.
If q+p is divisible by 12, it has to be divisble by 3 and 4. More importantly, if a number is divisible by 3 and by 4, it's divisible by 12 (if your teacher is a stickler for details, you may have to prove that also).
Let's make n equal to the smaller of p and q. That makes the other number equal to n+2. Let's first show that n + (n+2) is divisible by 4.
n+(n+2) = 2n + 2
=2(n+1)
We factored out a 2, so we know p+q is divisble by 2. Add since n is odd, n+1 is even and therefore also divisible by 2. So p+q is divisble by 4.
Now for divisble by 3. Just like in #4, for every (odd) number, one and only one of the following will be true:
Case 1: n mod 3 = 0;
Case 2: n mod 3 = 1; or,
Case 3: n mod 3 = 2
Keep in mind that n is a prime and also equal to the smaller of p and q. Since n is prime,
Case 1 can't be true. If Case 2 is true, then ((n+2) mod 3) = 0. That makes n+2 non-prime (divisble by 3). So Case 2 is also ruled out. That means Case 3 is true. Therefore,
((n+2) mod 3) = 1 and ((n + (n+2) mod 3) = 0. So, n + (n+2) is divisble by 3.
It now has been shown that n + (n+2) is divisble by 3 and by 4. Therefore it is also divisble by 12.
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Oh , right , just use modula to show can't all be primes .
Another question how to construct muplication table for ring ?
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