# SOLUTION: 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

Algebra ->  Algebra  -> Divisibility and Prime Numbers -> SOLUTION: 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       Log On

 Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Algebra: Divisibility and Prime Numbers Solvers Lessons Answers archive Quiz In Depth

 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