Proving $\frac{1}{2}(5x+4),\;2 < x,,\;\text{isPrime}(n)\Rightarrow n = 10k+7$ How is it possible to establish proof for the following statement?
$$n = \frac{1}{2}(5x+4),\;2<x,\;\text{isPrime}(n)\;\Rightarrow\;n=10k+7$$
Where $n,x,k$ are $\text{integers}$.

To be more verbose:
I conjecture that;
If $\frac{1}{2}(5x+4),\;2<x$ is a prime number, then $\frac{1}{2}(5x+4)=10k+7$
How could one prove this?
 A: If $n$ is prime , clearly $x$ must be even and moreover $x$ has only one power of 2, eg $x=2y$ where $y$ is odd, $y=2z+1$. Thus we have 
$$n\equiv 5y+2 \pmod{10}\equiv 5+2\equiv 7$$ 
A: What is $x$? I assume that it is some natural number: then the conjecture is false. If you let $x=0$, then $n=2$, but $n\equiv 2\text{ (mod }10)\not\equiv 7\text{ (mod 10})$.

For $n$ to be an integer, we must have $x$ even, and since $n\ne 2$, we must have $n$ odd. So in particular, we cannot have $5x+4$ even. This forces $x=4k+2$, hence
$$
\frac{1}{2}(5x+4)=\frac{1}{2}(5(4k+2)+4)=\frac{1}{2}(20k+14)=10k+7
$$
A: Since $n=5\,x/2\ +2$ is integer, we have that $x$ is even, and that $n$ gives remainder $2$ modulo $5$ (written $n\equiv 2\pmod{5}$). So, modulo $10$ there are only two possibilities: $2$ and $7=5+2$. As all primes $>2$ are odd, but $10k+2$ is even, it can't be prime.
(So, instead of 'prime' and $x>2$ you could have also said simply 'odd'.)
A: $x$ must be even or$n$ is not a natural. So let $x=2y$ and your conjecture is that if $5y+2$ is prime, it is$10k+7$. As $5$ divides into $10$, $5y+2 \equiv 2,7 \pmod {10}$. Any number $2\pmod {10}$ is even and (if $\gt 2)$ not prime.
