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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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    @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.2012-12-05

4 Answers 4

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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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    If I recall, this problem was on the AMC 10 test but I don't remember the year or the form. Either way, it is a very broad question that may appear anywhere else. Nice solution though.2016-08-11
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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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    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 $x + 2$.2012-12-04
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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 $\rm\:(x\!-\!y,\color{#0A0}y)=(x,y)=1.$