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.

  • 22
    $\begingroup$ The answer to 2 is certainly yes. While essentially the same proof, most would write (at most) something as short as this:"Assume $1007 = x+y$ with $x,y$ prime. As the sum of two odd numbers is even (and $1007$ is odd) one of $x,y$ must be even. Wlog. $x$ is even, and as $2$ is the only even prime we conclude $x=2$ and then $y=1007-2=1005=5\cdot 201$ is not prime, contradiction." - Then again, some folks would say that $-2$ is also a prime. In that case $1007=(-2)+1009$ is a solution! $\endgroup$ Nov 13, 2015 at 21:19
  • 2
    $\begingroup$ Goldbach's conjecture deals specifically with even integers. Why is this tagged with Goldbach's conjecture? $\endgroup$
    – austinian
    Nov 14, 2015 at 16:04
  • $\begingroup$ @austinian The question is loosely related to the conjecture. My professor says this question was meant as a thinking exercise as to why odd integers are not always the sum of two primes. $\endgroup$
    – klorzan
    Nov 14, 2015 at 20:25
  • 1
    $\begingroup$ Why number 1007, why not "007"? $\endgroup$
    – DVD
    Dec 23, 2015 at 4:51

4 Answers 4


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.

  • 8
    $\begingroup$ Well, yes, if you want to extend the concept of primality to the negative integers. But then you'd have to say that a prime number n is only divisible by 1, -1, n, and -n. Does anyone actually do this in serious mathematics? $\endgroup$
    – Tom Zych
    Nov 14, 2015 at 11:55
  • 7
    $\begingroup$ @TomZych Yes, in ring theory: en.wikipedia.org/wiki/Prime_element#Examples. $\endgroup$ Nov 14, 2015 at 13:29
  • 1
    $\begingroup$ This is why I miss studying math. Wonderful proof. $\endgroup$
    – Zaenille
    Nov 14, 2015 at 16:16
  • 1
    $\begingroup$ @TomZych it's necessary to think this generally if you want to talk about prime polynomials, or prime complex integers (a+bi). The integers is the unusual place we can chop off the negative numbers and have the theory work out nicely. $\endgroup$
    – djechlin
    Nov 14, 2015 at 18:16
  • 3
    $\begingroup$ For extra credit, show that there are no solutions in the Gaussian integers. :) $\endgroup$
    – PM 2Ring
    Nov 14, 2015 at 23:17

$1007$ is an odd number so it cannot be the sum of two odd numbers and it cannot be the sum of two even numbers. Therefore, it can only be the sum of an even and an odd number. Since $2$ is the only even prime it would have to be $2+1005$ and $1005$ is not prime.


HINT: $1007$ is odd. Suppose we have found a representation of $1007=x+y$ for $x$,$y$. Now, one of them is odd and the other even, or both odd. (Since $2$ is prime)


The claim is true. $1009$ and $-2$ are both prime numbers.

Edit: With apologies to @HagenvonEitzen the last sentence of whose comment to the OP I had not seen/parsed when I submitted this answer.

  • 3
    $\begingroup$ You can't be serious on that one!!! The term primality refers to natural numbers only. A number is prime if and only if it is divisible only by $1$ and by itself. Hence if you consider negative numbers as well, then there are no prime numbers at all!!! $\endgroup$ Nov 14, 2015 at 9:29
  • 2
    $\begingroup$ @barakmanos : No, there would still be prime numbers. An $n$ is prime if and only if $(n)$, the principal ideal generated by $n$, is a prime ideal. $-2$ qualifies (because $(-2) = (2)$ in $\Bbb{Z}$). The OP sets the ring with the word "integer" in "Lemma: An odd integer" and subsequent uses of "integers". $\endgroup$ Nov 14, 2015 at 9:38
  • 1
    $\begingroup$ @EricTowers Wiki says that prime element of a commutative ring is an object satisfying certain properties similar to the prime numbers, so your answer that -2 is a prime number is incorrect? $\endgroup$ Nov 14, 2015 at 18:28
  • 2
    $\begingroup$ This answer does not appear to address the poster's actual questions, which are about the quality of his proof and not about the veracity of the thing being proved. $\endgroup$ Nov 14, 2015 at 19:32
  • 3
    $\begingroup$ Small glitch in your comment above: $(0)$ is a prime ideal in $\mathbb{Z}$, but $0$ is not prime. $\endgroup$ Nov 15, 2015 at 9:02

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