187 – Quiz 2

Here is quiz 2.

Problem 1 asks to show that an integer is odd if and only if it is the sum of two consecutive integers.

Here is an argument: Let n be an integer. We need to prove two statements, corresponding to the two directions of the “if and only if:”

  1. If n is odd, then there are two consecutive integers a and b such that n=a+b.
  2. If n is the sum of two consecutive integers a and b, then n is odd.

1. Assume that n is odd. Then, by definition, there is an integer x such that n=2x+1. Note that 2x+1=x+(x+1), so we can take a=x, b=x+1, and we see that a and b are consecutive, and n=a+b.

2. Now assume that n=a+b where a and b are consecutive integers. Say, b=a+1. Then n=a+(a+1)=2a+1. By definition, this means that n is odd.

Problem 2 asks to show that if a,b,c are integers and we have that a|b and a|c then also a|(b+c).

To see this, assume that a,b,c are integers such that a|b and a|c. By definition, this means that there are integers x and y such that ax=b and ay=c. Then ax+ay=b+c. But ax+ay=a(x+y). Let z=x+y. Then z is an integer, and az=b+c. By definition, this means that a|(b+c).

Problem 3 asks to show that if n is a positive integer, then n^4-1 is composite.

As stated, this is false. For a counterexample, note that n=1 is a positive integer. However, 1^4-1=0 is not composite according to the definition given in the book: 

An integer a is composite if and only if there is an integer b such that 1<b<a and b|a.

In any case, it is true that if n is an integer and n>1, then n^4-1 is composite. To see this, note first that n^4-1=(n^2-1)(n^2+1). Now, if n>1, then n\ge2 and n^2\ge 4, so n^2-1\ge 3>1. Also, since 1<n^2-1, then 3\le n^2+1, and therefore n^2-1<3(n^2-1)\le (n^2-1)(n^2+1)=n^4-1. We have shown that (n^2-1)|(n^4-1) and that 1<n^2-1<n^4-1. By definition, this means that n^4-1 is composite.

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