Prove $x$ and $y$ in $y = x^2 + 2$ are prime only for $x = 3$ and $y = 11$?

Question

Let $x$ be a positive integer and $y = x^2 + 2$. Can $x$ and $y$ be both prime? The answer is yes, since for $x = 3$ we get $y = 11$, and both numbers are prime. Prove that this is the only value of x for which both x and y are prime.

I have to somehow prove this is true, and the only hint we were given is that we should consider cases depending on the remainder of $x$ modulo $3$. Does anyone have any tips on how to start proving this is true?

Answer 1

If $x\equiv 0\mod 3$, then it is obvious that for $x$ to be prime, $x$ must be $3$.

Now, take a look at what happens if $x\equiv 1 \mod 3$. What is the value of $x^2+2 \mod 3$? What about if $x\equiv 2\mod 3$?

Answer 2

first x and y are both odd positive integers, if x were even then x=2 is the only possible prime value then y would be 4+2=6 not a prime, also y can not be 2.

by division algorithm x can take 3 forms modulo 6 namely a- x≡1(mod 6) => x^2≡1(mod 6) b- x≡5(mod 6) => x^2≡25(mod 6) =>x^2≡1(mod 6) In either of the two previous cases y = x^2+2≡3(mod 6) ,then y is divisible by 3 for any value of x so y cant be prime . c-The last case when x≡3(mod 6)=> x^2≡9(mod 6) => y= x^2 + 2≡11(mod 6) Where the only possible prime value for x is 3 and its the corresponding y is 11 .