## 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 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 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 By definition, this means that $n^4-1$ is composite.