The claim is: 1007 can be written as the sum of two primes. We want to prove or disprove it.
Edit: My professor provided this definition in his previous assignment: An integer $n \geq 2$ is called prime if its only positive integer divisors are $1$ and $n$.
I want to disprove it.
Here is my proof outline:
Claim: 1007 can not be written as the sum of 2 primes.
Lemma: An odd integer can not be written as the sum of 2 even integers, or the sum of 2 odd integers. This means that an odd integer can only be written as the sum of an odd integer and an even integer.
Proof for lemma:
Let $a, b, c, d$ be integers.
$2a$ is even, $2b$ is even, $2c+1$ is odd, and $2d+1$ is odd.
$2a+2b=2(a+b)$ is even.
$(2c+1)+(2d+1)=2(c+d+1)$ is even.
$2a+(2c+1)= 2(a+c)+1$ is odd.
Thus, we have proved our lemma.
Since 1007 is odd, it can only be written as the sum of an odd integer and an even integer.
This means that if $x+y=1007$, for some integers $x,y$, then $x$ must be even and $y$ must be odd, without loss of generality.
We will show with cases that $x$ and $y$ can never both be prime.
2 is the only even prime number.
Case 1: $x=2$: $2+y=1007$, $y=1005$. Since 1005 is divisible by 5, it is not prime.
Case 2: $x$= any even integer $> 2$. According to our lemma, if $x$ is even, and $x+y=1007$, then $y$ must be odd. Every even integer greater than 2 is not prime, and so $x$ will always not be prime.
Thus, 1007 can not be written as the sum of two primes.
Thus we have disproved the original claim.
1) Is this proof complete?
2) Am I over complicating this?
3) Is there a more efficient way to prove this?
Any help would be appreciated.
Best Answer
Your work seems ok but too verbose. Here is a simple argument.
Suppose $1007=p+q$, with $p,q$ primes. Assume wlog that $p\le q$.
$p$ cannot be $2$ because then $q=1005$, which is not prime, being a multiple of $3$.
Therefore, $p\ge 3$ and so $q$ is also odd. But then $p+q$ is even and cannot be equal to $1007$, which is odd.