Lemon and melon

Let's take three positive integer numbers, a, b and c, so a+b, a+c and b+c are also integer numbers. We want a+b,b+c and c+a be prime numbers. If two of (a,b,c) are even, for example a and b, a+b is even. But the less value of a+b is 4, because the less value of a and b is 2 (2 is the smallest even number). And 4 isn't prime.

So there must be at least two odd numbers, for example a and b. But in this case a+b is even, so it must be 2, because we want a+b prime. Because of this, a and b must be 1. Let's think in a prime p. Now we know that a+c=b+c=1+c, so if a+c=p, c=p-1. And if instead of (a,b,c), we have (a',b',c'), (a',b',c')=(1,1,q-1), where q is another prime.

Now we have the only posibliity for (a,b,c): (1,1,p-1). Now we can say that (a,b,c) are the horizontal coordinates and (a',b',c') the vertical coordiantes of $\overrightarrow { u }$ , $\overrightarrow{ v }$ and $\overrightarrow { w }$ .

However, we have two 1 in each triple, so there are at least one vector, for example $\overrightarrow { u }$ , which must be (1,1). Also we must have a vector, for example $\overrightarrow { w }$ , which must be (1,wy). But wy cannot be 1, because in this way $\overrightarrow { u }$ and $\overrightarrow { w }$ will be equal, and it's forbidden for the problem.

So wy must be q-1. But in this way the angle of $\overrightarrow { u }$ is $\arctan { \frac { 1 }{ 1 } }$ and the angle of $\overrightarrow { w }$ is $\arctan { \frac { 1 }{ q-1 } }$ , so, if angle of $\overrightarrow { u }$ is the same as the angle of $\overrightarrow { w }$ , it means that 1= $\frac { 1 }{ q-1 }$ , or q-1=1. But it cannot be because $\overrightarrow { w }$ will be the same vector as $\overrightarrow { u }$ .

So, doesn't matter what are the coordinates of $\overrightarrow { u }$ , $\overrightarrow { v }$ and $\overrightarrow { w }$ , there isn't any three vectors which satisfy all the rules of the problem.

Thanks for reading and sorry for my english.