Question 4388
 You did a sloppy job in the posting without separating different questions and
 no marked numbers.

 Be careful about it.

  (i) if a and b are integers such that 6 divides both a+b and a-b, then 3 divides both a and b. 
 (ii) Show that the third power of every integer that is not divisible by 7 is of the form 7k+1 
 or 7k-1 where k is an integer. 
 (iii) Prove: if n is an integer, then one of the integers n, n+2, and n+4 is divisible by 3. 
 Use it to show that n=3 is the only integer such that n, n+2, and n+4 are primes. 
 (iv) Prove or give a counterexample: if d=(a,b), then 3d=(3a,3b) 
 (v) Prove or give a counterexample: if a and b are integers such that a^2 divides b^3, 
 then a divides b. 
 (v) Prove or give a counterexample: if the product of two integers is divisible 
 by 3, then at least one of the integers is divisible by 3.
 proof of (i) : Assume a+b = 6 k, a-b = 6j for some integer k,j
                then 2 a = 6(k+j) and 2 b = 6(k-j) 
                By cancelling 2,we have  a = 3(k+j) and  b = 3(k-j).
                This shows 3 divides both a and b. 
 proof of (ii) : if 7 is not a divisor of n, 
    (7k+1)^3 = 1 mod 7, (7k+2)^3 = 8 mod 7 = 1 mod 7,
    (7k+3)^3 = 27 mod 7 = 6 mod 7 = -1 mod 7,
    (7k+4)^3 = 4^3 mod 7 = 16* 4 mod 7 = 2*4 mod 7 = 1 mod 7,
    (7k+5)^3 = (-2)^3 mod 7 = -8  mod 7 = -1 mod 7 
    (7k+6)^3 = (-1)^3 mod 7 = -1 mod 7 
 (iii) Consider the three cases of n mod 3 below:
       when n = 0 mod 3, --> n+2 = 2 mod 3 and n + 4 = 4 mod 3 = 1 mod 3
       when n = 1 mod 3 -->  n+2 = 0 mod 3, n+4 = 5 mod 3  = 2 mod 3.
       when n = 2 mod 3 --> n+2 = 4 mod 3 = 1 mod 3, n+4 = 6 mod 3  = 0 mod 3.
      This shows among n, n+2, and n+4 , only one of them is divisible by 3. 
 
      Note 2 is the smallest prime,ie 1 is not a prime. 
      If n, n+2, and n+4 are primes(n>=2), since one of them is a multiple of 3. 
      We see that 3 is the only prime which is a multiple of 3, 
      But n+2 > 3, as a prime, we see that n+2 cannot be multiples of 3.
      Similarly n+4 cannot be multiples of 3.
      Hence, only n must ba multiple of 3 and so n = 3 .
      Note,n+2 =5, n+4 =7 are also primes.
      This shows n=3 is the only integer such that n, n+2, and n+4 are primes.      
  
 (iv) "if d=(a,b), then 3d=(3a,3b) "  is  True.
     Proof: d = (a,b)--> a = d k, b = dj for some integer k,j , &  (k,j) = 1
      --> 3a = 3dk, 3b=3dj for some integer k,j , &  (k,j) = 1 --> (3a,3b) = 3d 
 (v) "a^2 divides b^3 --> a divides b" is False
     For example: Set a= 2^3 = 8, b = 2^2 = 4, we have  a^2 | b^3 (ie 64 =8^2 | 4^3 = 64)
     But,  a | b ie 8| 4 is false

 (vi) In general, When p is prime then "p | ab --> p|a or p|  is True.
      Assume p is prime and p | ab.
      If p | a ie a = 0 mod p ,we are done.
      If a is not divisible by p, then (a,p) = 1, there exists integers n, m such that
      an + pm= 1 (Eucliden Algorithm) , this means an = 1 mod p.
     Since p |ab --> ab = 0 mod p --> nab = 1b = b mod p --> b is a multiple of p
     When the prime p = 3, we have 3 | ab  --> 3 | a or 3 |b.

 Try to read carefully about every step and hope that
 don't ask me what mod means.

 Kenny