Question 4388: I am having a lot of trouble with these proofs; I get so far and alskdjfl....
prove or give a counterexample: if a and b are integers such that 6 divides both a+b and a-b, then 3 divides both a and b.
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.
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.
Prove or give a counterexample: if d=(a,b), then 3d=(3a,3b)
Prove or give a counterexample: if a and b are integers such that a^2 divides b^3, then a divides b.
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.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! 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
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