$y^2=x^3+7$ has an integral soln. trivial issues that are imposing me from understanding a proof from a book. I have some trivial issues that are imposing me from understanding a proof from a book.
(i) if $y^2=8k^3+7$ where $k\in \mathbb{Z}$ $\implies$ $y^2\equiv7 \,[8]$ which is not possible, because $(\frac{7}{8})=-1$. But can someone tell me an easy way to interpret why 
$(\frac{7}{8})=-1$?
(ii) if $x$ is odd then I know why $(x-1)^2+3\equiv3 \,[4]$, but why does this imply that exists prime $p$ st $p|(x-1)^2+3\equiv3 \, [4]$ where $p\equiv3 \, [4]$? (according to book, otherwise all its prime factors are 1 modulo 4 $\implies$ $(x-1)^2+3\equiv1 \, [4]$) fair enough, but why does this $\implies$ $(x-1)^2+3\not\equiv3 \, [4]$? 
can a number not be congruent 1  [4] and 3  [4] simultaneously?
And does it have to be congruent to at least one of these?
(iii) $p\equiv3 \, [4]$ $\implies$  $(\frac{-1}{p})=-1$ ?
(iv) if $y^2=x^3+7$ has an integral soln. (which it does not) then $x$ has to be odd and $y$ even, why is $y$ necessarily even?
(v) $y^2=x^3+7$ reducing equation modolu $4$ $\implies$ $0\equiv(x+3) \,[4]$ I see what they have done, e.g. $7\equiv3 \, [4]$ so 7 becomes 3. 
how can we prove that $0\equiv(x+3) \, [4]$ is even true? 
(vi) $x$ odd and $x+2>0$ and $x+2\equiv3 \, [4]$ $\implies$ it has prime factor $p\equiv3 \, [4]$
why?
thanks a bunch

Here, I will try to be as explicit as possible to show that $y^2=x^3+7$ has no integer solution.
Case 1: Suppose that $x$ is even, so $x=2k$.  It follows that $y^2=8k^3+7$.  This implies that $y^2\equiv 7\mod 8$.  This is a contradiction because there is no $y$ such that $y^2\equiv 7\mod 8$.  You can check this by squaring the numbers $0,1,\ldots,7$, and showing that none of them give a remainder of $7$ when you divide by $8$.
Case 2: Suppose that $x$ is odd.  Notice that $x^3+7$ will be even, so that $y$ must also be even (else $y^2=x^3+7$ will be odd).  Note that $x^3\equiv x\mod 4$ (check this for $x=1,3$), $y^2\equiv 0\mod 4$ (because $y$ is even), and $7\equiv 3\mod 4$.  Thus, reducing our equation mod $4$ gives the equation $0\equiv (x+3)\mod 4$, or $x\equiv 1\mod 4$.
Now write $y^2+1=x^3+8=(x+2)(x^2-2x+4)$.  Since $x+2\equiv 3\mod 4$, there must exist some prime $p\equiv 3\mod 4$ dividing $x+2$.  This follows from the fact that all odd numbers are congruent to either $1$ or $3\mod 4$, and the product of two numbers congruent to $1\mod 4$ is also congruent to $1\mod 4$.  Then for this prime $p$ we can write $y^2+1\equiv 0\mod p$.  This is impossible by quadratic reciprocity.
Hence, no integer solutions exist.
