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Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Some problems from my textbook

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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Numbers are the essence of the Universe

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Some problems from my textbook

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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Numbers are the essence of the Universe

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: Some problems from my textbook

#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.

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: Some problems from my textbook

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.

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Some problems from my textbook

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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Numbers are the essence of the Universe