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Suppose that $x$, $y$, $x−y$, and $x+y$ are all positive prime numbers. What is the sum of the four numbers?

Well, I just guessed some values and I got the answer. $x=5$, $y=2$, $x-y=3$, $x+y=7$. All the numbers are prime and the answer is $17$. Suppose if the numbers were very big, I wouldn't have got the answer. Do you know any ways to find the answer?

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    $\begingroup$ I didn’t understand this question until I looked at the answers below. I thought the answer was “$3x+y$, obviously”. But now I see that you are looking for an actual number as an answer. The answers below show that there is only one set of values for $x$ and $y$ that makes the four integers positive and prime, so the actual work of the question is “what are the (only possible) values of $x$ and $y$?”. You’re saying you answered that question by brute-force guessing, but you want to know how to answer it with mathematical reasoning. $\endgroup$ – Rory O'Kane Dec 4 '12 at 21:50
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    $\begingroup$ The answers below also prove that there is a unique solution, which your guess gives you no reason to believe. $\endgroup$ – asmeurer Dec 4 '12 at 23:33
  • $\begingroup$ @RoryO'Kane Indeed, I had a hard time making sense of the question and answers until I realised the problem admitted only one solution which was implicitly required. $\endgroup$ – Thomas Dec 5 '12 at 9:11
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Note that $x>y$, since $x-y$ is positive. Since $x$ and $y$ are both prime, this means that $x$ must be greater than $2$ and therefore odd. If $y$ were odd, $x+y$ would be an even number greater than $2$ and hence not prime. Thus, $y$ must be even, i.e., $y=2$.

Now we want an odd prime $x$ such that $x-2$ and $x+2$ are both prime. In other words, we want three consecutive odd numbers that are all prime. But one of $x-2,x$, and $x+2$ is divisible by $3$, so in order to be a prime it must be $3$. Clearly that one must be $x-2$, the smallest of the three numbers, and we have our unique solution: $x=5$ and $y=2$, and $x+y+(x+y)+(x-y)=3x+y=17$.

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    $\begingroup$ The only reason it's the smallest of the three is that otherwise $x - 2$ is less than 2, the smallest prime. If the set were $\{x - 1, x, x + 4\}$ (for example), we could also conclude that one of the numbers must be 3, but in this case, the smallest element, $x - 1$, cannot be 3 (because then $x$ would be 4, which is not prime). $\endgroup$ – asmeurer Dec 4 '12 at 23:27
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    $\begingroup$ @asmeurer: Of course; I really thought that the reason was sufficiently obvious not to require explanation. $\endgroup$ – Brian M. Scott Dec 4 '12 at 23:32
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    $\begingroup$ Sure. It just sounds the way you've written it like you are saying it's $x - 2$ because that is the smallest. But I see now that you were just pointing out that that was the smallest, probably implicitly alluding to my argument above. $\endgroup$ – asmeurer Dec 4 '12 at 23:35
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    $\begingroup$ @IanOverton: He's not saying $x$ is divisible by three, but that one of {$x-2$,$x$,$x+2$} is divisible by three. That one must be three (since it's prime), and it must be the smallest (since three is the smallest odd prime), i.e. $x-2 = 3$. $\endgroup$ – hardmath Dec 5 '12 at 16:15
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    $\begingroup$ @Ian: As hardmath said, I’m not saying that $x$ is a multiple of $3$, just that one of the three numbers $x-2,x$, and $x+2$ has to be. You can verify this by brute force. If $x$ isn’t, then there is some integer $x$ such that $x=3n+1$ or $x=3n+2$. In the first case $x+2=3n+3$ is a multiple of $3$, and in the second case $x-2=3n$ is a multiple of $3$. Thus, in every case one of the three numbers is a multiple of $3$. Since these are primes, the one that’s a multiple of $3$ must be $3$. And since $3-2$ isn’t prime, $3$ must be the smallest of them, i.e., $x-2$. $\endgroup$ – Brian M. Scott Dec 5 '12 at 21:23
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Clearly $x>y$, since otherwise $x-y<0$ and therefore is not a prime.

Now since $x>y\ge2$, $x$ must be odd. Now $y$ must be even (i.e. 2) since if not $x+y$ is even and not prime.

The only set $\{x-2,x,x+2\}$ that consists of primes occurs when $x=5$. To see this, note that $\{x-2,x,x+2\}$ contains exactly $1$ number that is a multiple of $3$, so the multiple of $3$ in the set must be $3$ itself in order to be prime.

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Note that one of $x$ and $y$ has to be even, as if $x$ and $y$ are both odd, $x+y$ and $x-y$ are even, and there is only one even positive prime. As $x - y > 0$, we have $x > y$ and hence, as $2$ is the only even prime and the smallest prime, we have $y = 2$. No $x-2$, $x$ and $x+2$ are prime. But one of them is divisible by $3$: If $x$ has remainder 1 modulo 3, $x+2$ is divisible by 3, and if the remainder is 2, $x-2$ is. So, as $x-2$ is the smallest of the three numbers, we must have $x = 5$ (there is only one positive prime divisible by 3).

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    $\begingroup$ As with the other answer, there is no reason a prori to conclude that the smallest of the numbers must be 3. You either have to note that $x - 2$ would be less than 2 if $x$ or $x + 2$ were 3, or else just test all three possibilities ($x - 2=3$, $x=3$, and $x+2=3$) and note that only one of them gives three primes for $x - 2$, $x$, and $x + 2$. $\endgroup$ – asmeurer Dec 4 '12 at 23:32
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Hint $\rm \ \ x,\,y,\,x\pm y\,$ pair-coprime $\rm \iff x,y$ coprime, opposite parity, $ $ i.e. $\,\rm(x,y)=1=(x\!-\!y,2)$

Proof $\rm\ \, (x,x\pm y) = (x,y) = (x\pm y,y)\:$ so it reduces to: $ $ $\rm(x\!-\!y,x\!+\!y)=1$ $\!\iff\!$ $\rm\: (x\!-\!y,2)=1,\:$ assuming $\rm\:(x,y)=1.\:$ Then $\rm\ (x\!-\!y,x\!+\!y)=(x\!-\!y,\color{#C00}2\color{#0A0}y)=(x\!-\!y,\color{#C00}2),\ $ by $\rm\:(x\!-\!y,\color{#0A0}y)=(x,y)=1.$

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