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?

Mike Pierce
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    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. – Rory O'Kane Dec 04 '12 at 21:50
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    The answers below also prove that there is a unique solution, which your guess gives you no reason to believe. – asmeurer Dec 04 '12 at 23:33
  • @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. – Thomas Dec 05 '12 at 09:11
  • Some other questions with similar phrasing might allow multiple solutions for the variables, but all solutions share a single value for the requested computation. For example, "The side lengths of a right triangle are all natural numbers. One of the sides has length 5. What is the absolute difference between the other two side lengths?" has answer "1", though the triangle could be either $(3,4,5)$ or $(5,12,13)$. – aschepler Jan 09 '21 at 23:24

5 Answers5


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 $x+y+(x+y)+(x-y)=3x+y=17$.

Brian M. Scott
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    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). – asmeurer Dec 04 '12 at 23:27
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    @asmeurer: Of course; I really thought that the reason was sufficiently obvious not to require explanation. – Brian M. Scott Dec 04 '12 at 23:32
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    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. – asmeurer Dec 04 '12 at 23:35
  • Awesome question and answer to that awesome question. I still don't understand why x is divisible by 3. It looks like it goes up by 2 not 3....((x-2), x, (x+2)) -2, 0, +2. Would you please explain? – Ian Overton Dec 05 '12 at 14:36
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    @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. $x-2 = 3$. – hardmath Dec 05 '12 at 16:15
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    @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., $x-2$. – Brian M. Scott Dec 05 '12 at 21:23
  • 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. – Ian L Aug 11 '16 at 03:17

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 $x + 2$. – asmeurer Dec 04 '12 at 23:32

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 $\rm\:(x\!-\!y,\color{#0A0}y)=(x,y)=1.$

Bill Dubuque
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If $x,y > 1$ are both odd, then $x+y > 2$ is even, a contradiction.

Then $y = 2$ and by inspection $x=5$ as $x = 3$ results in $x-y = 1$.

Or, we have that $x-2,x,x+2$ are all prime meaning we have three consecutive integers modulo $3$ which is only satisfied by the triple $(3,5,7) \implies y = 5$.

Derek Luna
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