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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.

klorzan
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    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! – Hagen von Eitzen Nov 13 '15 at 21:19
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    Goldbach's conjecture deals specifically with even integers. Why is this tagged with Goldbach's conjecture? – austinian Nov 14 '15 at 16:04
  • @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. – klorzan Nov 14 '15 at 20:25
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    Why number 1007, why not "007"? – DVD Dec 23 '15 at 04:51

4 Answers4

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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.

lhf
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  • If you allow negative numbers, then $1007=(-2)+1009$, as pointed out by Hagen von Eitzen. – lhf Nov 14 '15 at 10:23
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    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? – Tom Zych Nov 14 '15 at 11:55
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    @TomZych Yes, in ring theory: https://en.wikipedia.org/wiki/Prime_element#Examples. – heinrich5991 Nov 14 '15 at 13:29
  • @heinrich5991: Interesting. Thanks. – Tom Zych Nov 14 '15 at 13:31
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    This is why I miss studying math. Wonderful proof. – Zaenille Nov 14 '15 at 16:16
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    @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. – djechlin Nov 14 '15 at 18:16
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    For extra credit, show that there are no solutions in the Gaussian integers. :) – PM 2Ring Nov 14 '15 at 23:17
  • Why don't you have to prove that two odd numbers summed can't be even? – Tony Ennis Nov 15 '15 at 01:44
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    @TonyEnnis: 1). 1007 is odd, so we only need to discuss pairs of numbers that sum to an odd number. 2). It's never a good idea to prove things that are false. :) – PM 2Ring Nov 15 '15 at 01:58
  • @PM2Ring I get that. But this proof assumes two odds can't sum to an odd. While it is true, why doesn't it have to be proven? – Tony Ennis Nov 15 '15 at 02:00
  • @TonyEnnis: Fair point. Strictly speaking, you should prove that odd + odd must be even, rather than simply assuming it. OTOH, if you prove every trivial thing in a proof you'll soon get bogged down in minutiae, so in any mathematical field it's customary to assume certain core results that anyone working in the field should be familiar with. – PM 2Ring Nov 15 '15 at 03:21
  • (cont) So when doing proofs about primes it's reasonable to assume the fundamental properties of addition & multiplication of odds & evens. Or at least, to merely give a reference to the proofs of such properties, rather than being obliged to prove them explicitly. – PM 2Ring Nov 15 '15 at 03:21
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$1007$ 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.

John Douma
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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)

9301293
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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.

Eric Towers
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    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!!! – barak manos Nov 14 '15 at 09:29
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    @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". – Eric Towers Nov 14 '15 at 09:38
  • @barakmanos : See, for instance, https://en.wikipedia.org/wiki/Prime_element , where you learn "The integers ±2, ±3, ±5, ±7, ±11, ..." are all prime in $\Bbb{Z}$. – Eric Towers Nov 14 '15 at 09:39
  • -2 is not a prime number in this case. My professor has given us the definition of prime to only include positive integers. – klorzan Nov 14 '15 at 10:58
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    @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? – Peeyush Kushwaha Nov 14 '15 at 18:28
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    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. – Daniel Wagner Nov 14 '15 at 19:32
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    @PeeyushKushwaha : So, what? It's a prime frog? A prime pump? Claiming that $-2$ is not a number is substantially more bizarre than recognizing that it is a prime. – Eric Towers Nov 14 '15 at 22:23
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    Small glitch in your comment above: $(0)$ is a prime ideal in $\mathbb{Z}$, but $0$ is not prime. – Daniel Fischer Nov 15 '15 at 09:02
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    @EricTowers English might work that way in general, but that's only if definitions are used in their plain English meaning. "prime" can mean either "prime number" or "prime element". You're using it as "prime element". In that case, "x is a prime and x is a number" does not imply that "x is a prime number", because "prime number" has a separate definition that excludes some things. Admittedly, the OP uses "prime" without adding "number" for the most part, which would allow it to mean "prime element", but does use "prime number" in one place, making it clear which meaning of "prime" is used. – hvd Nov 15 '15 at 10:53
  • @DanielFischer : You are correct, sir. I would edit my comment above, but that's not really a thing here. – Eric Towers Nov 17 '15 at 02:22
  • @hvd: One could alternatively observe that $\Bbb{Z}$ is the ring of integers of the algebraic number field $\Bbb{Q}$, so all of its elements are numbers. However, none of this semantic hair-splitting resolves the underlying problem: The OP never explicitly specifies what ring he's working in. His repeated use of the word "integer" is the strongest hint. His single use of the word "number", buried in his proof, need not be normative. – Eric Towers Nov 17 '15 at 02:33
  • @EricTowers This is not a proof of falsehood. My professor provided this definition of prime:

    Def: An integer $n \geq 2$ is called prime if its only positive integer divisors are $1$ and $n$.

     – klorzan Nov 20 '15 at 00:55
  • @klorzan : Excellent. Where is that definition in your post? – Eric Towers Nov 20 '15 at 03:29