If $x$, $y$, $x+y$, and $x-y$ are prime numbers, what is their sum?

Question

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?

Answer 1

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

Answer 2

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).

Answer 3

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.

Answer 4

Hint $\rm \,\ x,\,y,\,x\pm y\,$ pair-coprime $\rm \!\iff\! x,y$ coprime, unequal 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 $\rm\:(x\!-\!y,\color{#0A0}y)=(x,y)\!=\!1.$

Answer 5

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 $(3,5,7) \implies y = 5$.